A digestion of the proof of Sendov's conjecture

# A digestion of the proof of Sendov’s conjecture Source: [https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/) This post concerns the following conjecture[of Sendov](https://zbmath.org/3254142), as well as its strengthening by[Phelps–Rodriguez](https://zbmath.org/3386011): > **Conjecture 1 \(Sendov’s conjecture\)**Let, and letbe a degreepolynomial with all zeroes in the unit disk\. Then for every zeroof, there exists a critical pointofwith\. > **Conjecture 2 \(Phelps–Rodriguez conjecture\)**Let, and letbe a degreepolynomial with all zeroes in the unit disk\. Then for every zeroof, there exists a critical pointofwith, unlessis on the unit circle andis a scalar multiple of\. By applying a rotation around the origin, we can normalizeto be a real number with\. From the[work of Rubinstein](https://zbmath.org/3274887), both conjectures were already established in thecase, so one can restrict to thecase\. Both of these conjectures then follow from > **Conjecture 3 \(Sendov’s conjecture in interior\)**Let\. Letbe a degreepolynomial with all zeroes in the unit disk\. Then ifis a zero of, there exists a critical pointofwith\. All three of these conjectures were established for\(in a sequence of papers culminating in[this paper of Brown and Xiang](https://zbmath.org/1286149)\) and for sufficiently large\(in[a paper of myself](https://terrytao.wordpress.com/2020/12/08/sendovs-conjecture-for-sufficiently-high-degree-polynomials/), which in turn built upon several partial results in this setting\)\. This left the case of intermediateto be settled\. My arguments used some qualitative ingredients \(most notably analytic continuation\) and as such did not easily lend themselves to quantifying the threshold ofabove which the argument was valid\. Recently, Lech Mazur[was able to use an AI tool](https://www.proofatlas.ai/papers/sendov-conjecture/SENDOV_CONJECTURE_PROOF_AUGUST_5_2026.pdf)to resolve Sendov’s conjecture for all, with the proof[verified in Lean](https://www.proofatlas.ai/formalizations/sendov-conjecture/)\. However, the AI\-generated proof was not human\-digested to be in the form of a publication\-ready preprint; and it has taken me several days \(with[heavy AI assistance](https://chatgpt.com/share/6a7ceea8-5aa0-83e8-a56b-67cd7ec59f06)\) to perform such a digestion, to place the proof in proper context with previous literature and to simplify and streamline the argument to highlight the main ideas\. \(Note: the above chat log only represents a portion of the digestion work: the rest was performed with pen and paper, or using some further AI agents\.\) The same arguments also give a new proof of Rubinstein’s theorem, which I also give below the fold\. One consequence of this digestion is that the argument in fact demonstrates Conjecture[3](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#interior), and thus resolves both the Sendov conjecture and the Phelps–Rodriguez conjecture in full generality\. The proof ends up being remarkably elementary\. No complex analysis is used other than the fundamental theorem of algebra \(and very basic facts about Möbius transformations\); and the deepest inequality used as input is the[Maclaurin inequality](https://en.wikipedia.org/wiki/Maclaurin%27s_inequality)\(and we only need a special case of that inequality which can be derived from the arithmetic mean\-harmonic mean inequality and an induction argument\)\. Using an AI agent, I[have been able to formalize](https://github.com/teorth/sendov)the entire argument in Lean, extended toby some minor modifications to the proof\. This formalization is more streamlined than the original formalization \(it has about 15,000 lines of code, compared with around 90,000 for the original proof\)\. We now prove Conjecture[3](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#interior)\. Thecases have long been known but need to be treated separately; a short proof using the machinery developed here is provided at the end of the post\. Suppose now that we have a counterexample for some, thus one can find a degreepolynomialwith zeroes  for someand,in the closed unit disk, whose critical points all lie a distance at leastfrom\. We usenotation here in the non\-asymptotic sense, thusmeans thatfor some absolute constant\(independent of\)\. We will also use the notationto denote a quantity that is bounded in magnitude by\. To capture the fact that the critical points lie at a distance at leastfrom, we write these critical points as  for some \(non\-zero\)in the closed unit disk\. > **Example 4**Ifand, thenare the non\-trivialroots of unity, while theare all equal to\. Strictly speaking this is not actually a counterexample to Conjecture[3](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#interior), becauseis not strictly less than one; nevertheless this is an important motivating near\-counterexample for the arguments below\. > **Example 5**A generalization of the previous example was studied in Section 4 of[my paper](https://arxiv.org/abs/2012.04125)\. Here one took wherewas an asymptotic parameter going to infinity, was a low\-degree polynomial for some, andwere constants\. This polynomial has a zero at,critical points at, andadditional critical points near\. If all the critical points were at distance at least one from, one would have and while if all the zeroes were in the unit disk, the calculations in my paper showed thatHeredenotes a quantity that goes to zero as\. If one ignores theerrors, one can show that these conditions are only simultaneously feasible ifand all thevanish, but the argument was somewhat subtle \(I had to proceed by inspecting the second Fourier coefficient of[\(1\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#c21)\)\. This illustrates the fact that the regimeis particularly delicate\. We now have two sets of points in the closed unit disk:and\. They “communicate” with each other through the polynomialand its first derivative, both of which can be expressed in terms of either set of points \(as well asand\)\. Indeed, if we normalizeto be monic, then we can factorin terms of the zeroes as  and thus upon differentiating Here and in the sequel we adopt the convention of removing singularities when dealing with expressions that involve multiplication by bothand, by cancelling such terms first in the event that\. In a similar vein,can be factored  and thus on integrating \(and using\) It is convenient to rule out the easy caseright away\. In this case we see from[\(3\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#p-deriv-zero),[\(4\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#p-factor)that  which is absurd since the first product has magnitude at most one, and the second product has magnitude at least one\. Thus we can assume henceforth that\. By inspectingorat various natural locations, we can thus obtain a number of identities relating theto the\. We record the ones that we actually need here: > **Lemma 6 \(Communication identities\)**Letdenote the function- \(i\) \(Centroid identity\) We haveThat is to say, the centroid of the zeroes equals the centroid of the critical values\. - \(ii\) \(Polar identity\) We have - \(iii\) \(First origin identity\) We have - \(iv\) \(Second origin identity\) We have\(Again, we are using the convention of removing singularities to deal with the case where some of thevanish\.\) *Proof:*For \(i\), we inspect the behavior ofas\. From[\(2\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#p-eq-zero)we have  and thus on differentiating term by term  Meanwhile, from[\(4\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#p-factor)we have  Comparing coefficients, we obtain the claim\. For \(ii\), we consider the expression\. On the one hand, from[\(2\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#p-eq-zero),[\(3\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#p-deriv-zero)one has  \(Note from hypothesis thatcannot be a critical point, so the denominator is non\-zero\.\) On the other hand, from[\(4\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#p-factor),[\(5\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#p-integral)one has  Equating the two identities, we obtain \(ii\) after some algebra\. For \(iii\), we evaluate\. From[\(2\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#p-eq-zero)we have  while from[\(5\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#p-integral)we have  Equating the two identities, we obtain \(iii\) after some algebra using[\(6\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#F-def)\. For \(iv\), we similarly evaluate\. From[\(3\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#p-deriv-zero)we have  while from[\(4\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#p-factor)one has  Equating the two identities, we obtain \(iv\) after some algebra using[\(6\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#F-def)\. Remarkably, the polynomialwill play no further role in the argument: the identities in \(i\)\-\(iv\), together with the hypotheses thatandlie in the closed unit disk, will be sufficient by themselves to obtain a contradiction\. > **Example 7**Continuing the example in Example[4](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#ex), in \(i\) both sides vanish\. In \(ii\), both sides are equal to one\. For \(iii\) and \(iv\), we have, with both sides of \(iii\) equal to one, and both sides of \(iv\) equal to zero\. > **Remark 8**The centroid identity is extremely classical, going back to this 1948[paper of Popoviciu](https://zbmath.org/3047862)\. The comparison of the polynomial at a locationand at the polar inversionof that location across the closed unit disk is a familiar trick in the literature; see, e\.g\., Lemma 5 and Theorem 8 of[Dégot](https://zbmath.org/6269492)\. The specific form of the polar identity is implicit in the first part of Section 5 of[Mazur’s AI\-generated proof](https://www.proofatlas.ai/papers/sendov-conjecture/SENDOV_CONJECTURE_PROOF_AUGUST_5_2026.pdf), while the origin identities are extracted from equation \(6\.3\) of that proof\. The first origin identity is also very close to Theorem 6 of Dégot, while the second origin identity is similar to some identities appearing in the proof of Lemma 6 of Dégot, as well as the work of[Mir–Nazir–Wani](https://zbmath.org/7742078)and \(in thecase\)[Rubinstein](https://zbmath.org/3274887)\. The work of[Meir–Sharma](https://zbmath.org/3308557)and[Mir–Nazir–Wani](https://zbmath.org/7742078)also contain several further identities relating theto the; see in particular Lemma[15](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#meir)below\. Variants of[\(5\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#p-integral)also appear in Proposition 10 of[Miller](https://zbmath.org/399047)\. > **Remark 9**The first origin identity[\(9\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#first-origin)is already strong enough to handle asymptotically all examples of the form in Example[5](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#ex2), except in the endpoint case wherevanish and theare all\. Indeed, as theare in the closed unit disk,[\(9\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#first-origin)implies that On the other hand, routine calculations \(omitted here\) show that leading asymptotically to the constraint But all terms here are non\-negative \(since\), so this forces a contradiction unless\(and hence also\) and theall vanish\. As mentioned in Example[5](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#ex2), the most delicate regime occurs when\. It is convenient to introduce the normalized version  of, thus, and the casecorresponds to\. Informally,measures how closeis to\(at the scale of\)\. A key role in the argument will be played by the mean  of the, particularly the real part\. As theall lie in the unit disk, the meandoes also, so that  and On the other hand, in the example in Example[4](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#ex),is equal to the extremal value of, and\. In Example[5](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#ex2), we have\(and\)\. It will be convenient to work with the quadratic polynomial  with a particular emphasis on the value at: One should primarily think ofas a measure of how closeis to\. Clearly we have  for all\(note thatis strictly less than\)\. The arguments will revolve around the relationship betweenand\. Specifically, we will establish the following two inequalities below the fold\. The first inequality, which we call the “polar inequality”, comes in three forms: > **Proposition 10 \(Polar inequality\)**- \(i\) \(Raw polar inequality\) We have - \(ii\) \(Polar inequality in,form\) We have - \(iii\) \(Simplified polar inequality\) We haveIn particular, since, one has It will be the inequality[\(18\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#beta-bound)that we use in practice, but it will be derived from[\(17\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#lt), which in turn is a consequence of[\(16\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#1Q), which will follow from the polar identity[\(8\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#polar-ident)together with the fact that theandlie in the unit disk\. The bound[\(18\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#beta-bound)is only slightly weaker than[\(17\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#lt); see the \(Gemini\-generated\) image below\.  I was not able to find an exact duplicate of the above polar inequalities in past literature, but the[paper of Dégot](https://zbmath.org/6269492)contains several similar inequalities\. The inequality[\(16\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#1Q)was extracted from \(5\.1\) of[Mazur’s AI\-generated proof](https://www.proofatlas.ai/papers/sendov-conjecture/SENDOV_CONJECTURE_PROOF_AUGUST_5_2026.pdf); the subsequent bounds[\(17\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#lt),[\(18\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#beta-bound)arose from my attempts to simplify the arguments after that point\. The second inequality, which is more difficult, also will come in several forms: > **Proposition 11 \(Origin inequality\)**Let\.- \(i\) \(Raw origin inequality\) We have - \(ii\) \(bound\) We have - \(iii\) \(Origin inequality in,form\) We have Part \(i\) \(which was extracted with some effort from Section 6 of the[original AI\-generated argument](https://www.proofatlas.ai/papers/sendov-conjecture/SENDOV_CONJECTURE_PROOF_AUGUST_5_2026.pdf)\) will be deduced from the first and second origin identities[\(9\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#first-origin),[\(10\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#second-origin), as well as the centroid identity[\(7\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#centroid-ident)\. Part \(ii\) will follow from \(i\) and the polar inequality[\(18\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#beta-bound), while part \(iii\) is an elementary consequence of \(i\)\. As it turns out, the last three terms in[\(21\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#1le)are asymptotically negligible as\. Dropping those terms gives a competing feasibility region forandwhich is disjoint from the one coming from the polar inequality[\(17\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#lt)\(or[\(18\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#beta-bound)\):  This already suggests that one can use this approach to recover my previous result on Sendov’s conjecture holding for all sufficiently large\. In fact, even with the three error terms in[\(21\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#1le)added, there is enough room between the two inequalities[\(18\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#beta-bound),[\(21\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#1le)to obtain a contradiction for all\(using the additional boundto control these errors\), although showing this for medium\-sized\(such as\) requires a certain amount of computer assistance\. For fixed, the right\-hand side of[\(21\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#1le)is monotone increasing in\(or equivalently, monotone decreasing in\)\. In view of[\(18\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#beta-bound), we can thus replacebyin this inequality, so thatis replaced by  andreplaced by\. The inequality[\(21\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#1le)then becomes an inequality involving onlyand: We also note that the boundsforce the constraint  This preventsfrom getting too close to the upper limit\(orgetting too close to zero\)\. We can now eliminate all large degrees, e\.g\.,, as follows\. The quadraticattains its minimum at\. Forwe have  while for\(if this region is non\-vacuous\) we can bound the quadratic by its valueat\. Thus  Evaluating these expressions, we arrive at  Since, we have\. Next, we claim that\. Asis monotone increasing in, it suffices to do this when\. Here one can directly compute that  since the discriminantof the numerator is negative, we conclude that  as desired\. Dropping someandterms, we conclude that  Every term on the right\-hand side can be seen to be decreasing infor\. Thus the right\-hand side can be bounded by  giving the desired contradiction\. The remaining range to handle is when  It turns out that[\(22\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#stat)remains infeasible in this range\. This can be illustrated numerically without much difficulty: see[this applet](https://teorth.github.io/tao-web/apps/sendov.html)\. For instance, in the most delicate case, the right\-hand side of[\(22\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#stat)only gets as large as\(and in particular stays below\) throughout the range:  I have also[verified this bound in Lean](https://github.com/teorth/sendov)\. **— 1\. The polar inequality —** We begin with a proof of Proposition[10](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#polar)\. As is well known, the Möbius transformmaps the closed unit disk to itself\. In particular, we have  for all of the zeroes\. Inserting this into the polar identity[\(8\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#polar-ident)and using the triangle inequality, we conclude the lower bound We now convert this bound to a bound involving the quantityin[\(12\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#xy-def)\. From the arithmetic mean\-geometric mean inequality we have and from[\(12\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#xy-def)we have  Since, we thus have giving the raw polar inequality[\(16\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#1Q)\. Boundingbyand using the quantitiesfrom[\(11\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#alpha-def),[\(15\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#beta-def), we observe that  Using the basic inequality, we thus have  with strict inequality for\. From[\(16\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#1Q)we conclude[\(17\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#lt)\. This also implies, since otherwise the integrand is always bounded by, which is absurd\. On evaluating the integral in[\(17\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#lt), we obtain  and thus  so on taking logarithms we obtain  It remains to establish the bound Here we use an AI\-generated argument\. One can directly calculate  whereand\. If we can show that for all, then taking logarithms in[\(17\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#lt)yields  from which[\(26\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#polar-corollary)will follow by routine algebra\. Both sides of[\(27\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#lsh)vanish at\. Taking derivatives, it suffices to show that  which rearranges to  To expand the left\-hand side, we use the double angle formulaeand  to rewrite it as  Collecting the coefficient offorand extracting a common factor of, one is left with  where\. \(The remaining coefficients, which also receive contributions from the polynomial terms, all vanish\.\) Thus the left\-hand side has the Taylor expansion  in which every coefficient is non\-negative, giving the claim\. > **Remark 12**As the image in the introduction suggests, the bound[\(18\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#beta-bound)is only slightly weaker than[\(17\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#lt)\. For small, one can perform Taylor approximation on the latter bound to obtain while the former bound is Note thatis slightly smaller than\.Relating to this, the constantin[\(27\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#lsh)cannot be improved\. **— 2\. The origin inequality —** Now we turn to the proof of Proposition[11](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#origin), which is more difficult and revolves around an analysis of the functiondefined in[\(6\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#F-def)\. We begin with a heuristic analysis\. Inserting the approximationfor smallinto[\(6\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#F-def)and using[\(12\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#xy-def), we are led to the approximation  at least whenis small \(which turns out to be the dominant regime in applications\)\. This suggests a relation between the two expressions involvingin the origin identities in Lemma[6](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#identities)\. Substituting in this approximation, we obtain some \(slightly complicated\) approximation for the sumin terms of,,,, and the product\. Aslie in the closed unit disk, the productdoes also\. However, past experience with the Sendov conjecture has taught us that the worst cases tend to be whenlie very close to the boundary of the disk, so thatis close to one\. For instance, in Example[4](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#ex)all theandlie on the unit circle, and\. See Remark 3 of[Dégot](https://zbmath.org/6269492)or Theorem 1\.10\(ii\) of[my own paper](https://zbmath.org/7681950)for other places where this heuristic is noted\. To simplify the discussion, let us assume for now thatis exactly one, so thatall lie on the unit circle\. This leads in particular to the inversion identities  The centroid identity in Lemma[6](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#identities)\(i\) relates the sum of thewith the sum of the\. Using[\(31\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#inverse), this gives a similar identity relating the sum of thewith the sum of the\. The latter sum is of course just\. This combines well with the previous approximation, thus giving an approximate identity relating,to,, and\. As it turns out, the roles ofandare minor and can be quickly eliminated for the purposes of obtaining useful bounds, leading eventually to the relation in Proposition[11](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#origin)\. We turn to the details\. To make the approximation[\(30\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#1fq)more precise, we note that, and hence by the fundamental theorem of calculus  The heuristic[\(29\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#ft)predicts that, which would give[\(30\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#1fq)\. If we actually differentiate[\(6\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#F-def)carefully, we obtain the exact identity  Bounding, we write this  When faced with a similar expression in[\(24\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#amgm), we used the arithmetic mean\-geometric mean inequality\. Here, the analogous tool is[Maclaurin’s inequality](https://en.wikipedia.org/wiki/Maclaurin%27s_inequality), which gives  and hence by Cauchy–Schwarz  Repeating the calculations used to show[\(25\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#quad), we have  and so we obtain the bound  Integrating this, we obtain a rigorous analogue of[\(30\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#1fq),  and thus by the triangle inequality From the first and second origin identities[\(9\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#first-origin),[\(10\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#second-origin)we have The next step is thus to estimate\. When, then all thewere on the unit circle and we could use[\(31\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#inverse)\(and the centroid identity\) to proceed\. Now, we are no longer assumingto equal, but we can still adapt the previous arguments with a loss proportional to\. The key lemma is > **Lemma 13 \(Defect lemma\)**Letbe some points in the closed unit disk\. Then *Proof:*By a limiting argument we may assume that none of thevanish\. If we writefor some, then we can calculate that  and  Thus the desired inequality reduces to the superadditivity property  But from the sinh addition formulawe have  for all non\-negative\(this also follows from the convex nature oftogether with\), and the claim follows by induction\. We remark that the lemma can also be proven by direct induction, without an appeal to hyperbolic trigonometry\. From taking complex conjugates of the centroid identity[\(7\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#centroid-ident)and performing some algebra, we have  Using the defect lemma \(applied to the points\) and the triangle inequality we conclude that  where as before we are removing singularities when some of thevanish\. Applying[\(12\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#xy-def)and some algebraic manipulation, we arrive at  Substituting this back into[\(33\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#f1aq), we conclude that  and hence after some algebra and the triangle inequality  Inserting this into[\(32\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#tri), we obtain We can simplify[\(34\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#tri-2)by reducing to thecase\. Indeed, we shall show that which implies that the right\-hand side of[\(34\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#tri-2)is non\-decreasing inin the range\. Thus we may replacebyin[\(34\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#tri-2)to conclude that Let us now verify[\(35\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#grow)\. Usingand the triangle inequality, we can lower bound  Inserting this into[\(35\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#grow)and clearing denominators, we reduce after some algebra to  But as a quadratic polynomial in, the left\-hand side has discriminant, which one can check to be negative for sufficiently large\(in factsuffices\), giving the claim[\(35\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#grow)\. Next we eliminate the role of the imaginary term\. Observe for any complex numberwith positive real part that  as can be seen by squaring both sides\. The expressionhas real part  which lies betweenand\(in particular, it is positive\), and imaginary part of magnitude at most  by[\(13\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#ybound)\. We conclude that  The right\-hand side can be rearranged using the quantityfrom[\(11\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#alpha-def)as  so the bound[\(36\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#tri-3)gives[\(19\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#origin-exact)\. **— 2\.1\. Upper bound on—** Now we can prove[\(20\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#17)\. Suppose for contradiction that; since, this implies that\. Crudely discarding theterm in[\(19\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#origin-exact)and boundingby, we have  The quadratic polynomialequalsatand attains its minimum atwith value\. By convexity, we thus have  forand  for\(this latter statement is vacuous if\)\. Since, we can therefore crudely bound   and hence  From[\(15\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#beta-def)we have, thus by[\(18\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#beta-bound)one has  From another application of[\(18\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#beta-bound)one has  We conclude that  It is now convenient to introduce the quantity, thuswith  and  Inserting these bounds and dividing by, we conclude  Since, we obtain  Sinceand, we conclude that  Routine calculus shows thathas a maximum of at most, and that the right\-hand side here is at most, giving the required contradiction\. This proves[\(20\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#17)\. **— 2\.2\. A simplified estimate —** Now we show[\(21\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#1le)\. Note from[\(11\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#alpha-def)that  while from[\(15\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#beta-def)we have and hence also From[\(15\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#beta-def)we have  By the mean value theorem \(noting thatis non\-negative\) we thus have  From the standard beta function identity  \(and the fact that\) we can thus replace[\(19\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#origin-exact)by  From[\(15\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#beta-def)we have  Thus by[\(37\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#1a),[\(38\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#1x),[\(39\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#1x')  Dividing by the positive quantitygives the claim\. **— 3\. Rubinstein’s theorem —** We now adapt the arguments to give a proof of Rubinstein’s theorem that the Phelps–Rodriguez conjecture holds in thecase, i\.e\., > **Theorem 14 \(Rubinstein’s theorem\)**Let, and letbe a degreepolynomial with all zeroes in the unit disk\. If, then there exists a critical pointofwith, unlessis a scalar multiple of\. The argument here is essentially in Remark 5\.1 of[this paper of Tang and Zhang](https://arxiv.org/abs/2508.10341)\. Taking contrapositives, we may assume that the critical pointsare of the formfor somein the closed unit disk, and normalizeto be monic; our task is to show that\. The polar identity[\(8\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#polar-ident), based on calculatingdegenerates to a triviality when, but we have the following usable substitute, valid for any choice of, first observed in equation \(3\.2\) of[Meir–Sharma](https://zbmath.org/3308557): > **Lemma 15 \(Meir–Sharma identity\)**Ifand the critical points are of the formthen all the zeroesare not equal to, and *Proof:*By hypothesis,is not a critical point of, soandfor all\. Instead of computing, we instead consider the expression\. On the one hand, from[\(4\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#p-factor)we have  while from differentiating[\(4\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#p-factor)we have  Meanwhile, from[\(3\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#p-deriv-zero)we have  and from differentiating[\(3\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#p-deriv-zero)we have  Using these identities to computein two different ways gives the claim\. Now take\. Sincelie in the closed unit disk,has real part at least, whileis at most\. Thus, the only way that the above identity can hold is iffor all, hencefor all\. Thus all critical points are at the origin, which forcesfor some\. Since, we conclude that, giving the claim\. **— 4\. Thecases —** We now prove thecases of Conjecture[3](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#interior)\. The starting point is[\(23\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#polar-lower-bound)\. Using the triangle inequality and, this implies that  \(This also follows from[\(16\)](https://terrytao.wordpress.com/2026/08/12/a-digestion-of-the-proof-of-sendovs-conjecture/#1Q)and\.\) From Hölder’s inequality andwe conclude that  The right\-hand side can be computed to equal  which is obviously less thanfor, giving the contradiction\. > **Remark 16**The same argument also works for, but breaks down for higher\. **— 5\. Further directions —** The Sendov and Phelps–Rodriguez conjectures are now resolved, but several related conjectures remain open\. The following strengthening of Sendov’s conjecture, by Borcea, is open for any: > **Conjecture 17 \(Borcea conjecture\)**Letand, and letbe a degreepolynomial with zeroessatisfying\. Then for every zeroof, there exists a critical pointofwith\. Sendov’s conjecture is the limiting caseof this conjecture\. There has been relatively little progress on this conjecture: the caseswere established[by Khavinson, Pereira, Putinar, Saff, and Shimorin](https://zbmath.org/6075447), and in[this previous paper](https://terrytao.wordpress.com/2025/11/05/mathematical-exploration-and-discovery-at-scale/)we reported the negative result that AlphaEvolve failed to find a counterexample to the conjecture\. The proof methods here do not seem to extend easily; all the identities relating zeroes and critical points continue to hold, but now that theare only constrained to the unit disk in an averaged moment sense, all of the inequalities developed above now fail\. Another strengthening of Sendov’s conjecture that remains open is[Schmeisser’s conjecture](https://zbmath.org/3539507): > **Conjecture 18 \(Schmeisser’s conjecture\)**Let, and letbe a degreepolynomial with all zeroes in the closed unit disk\. Then for anyin the convex hull of the zeroes of, there exists a critical pointofwith\. Schmeisser proved several special cases of this conjecture, and AlphaEvolve again failed to find a counterexample, but there has not been much further progress\. Here, theare now back in the closed unit disk, but we no longer have, again rendering most of the previous identities invalid\. But perhaps some modification of the arguments here can make some progress on this conjecture\. A common generalization of the Borcea and Schmeisser conjectures was proposed in Conjecture 2\.4 of[this paper of Zhang](https://arxiv.org/abs/2411.07105)\. A slightly different strengthening was also proposed in Conjecture 1\.10 of[Tang and Zhang](https://arxiv.org/abs/2508.10341): > **Conjecture 19 \(Tang–Zhang conjecture\)**Let, and letbe a degreepolynomial with all zeroes in the closed unit disk and critical points\. Then for any, one has\. Sendov’s conjecture is the limiting case\. By Hölder’s inequality, thecase is the strongest form of the conjecture\. Another well known variant of Sendov’s conjecture is[Smale’s problem](https://zbmath.org/3713836): > **Conjecture 20 \(Smale’s problem\)**Let, and letbe a degreepolynomial\. Then for any zeroof, there exists a critical pointofwith\. The constantis best possible, as can be seen by the exampleand\. Using the Koebe one\-quarter theorem, Smale proved this conjecture withreplaced by\. Some slight improvements of this bound have been obtained over the years; for instance for, the improved bound ofwas[obtained by Crane](https://zbmath.org/5214827)\. Again, AlphaEvolve failed to find a counterexample to this conjecture\. This problem does not seem to have a direct relationship with Sendov’s conjecture, and there is no useful normalization of the zeroes and critical points that is confined to the unit disk\. Nevertheless there may be some hope of making progress on this conjecture, perhaps working first in the asymptotic regime\. Needless to say, I did try some desultory attempts to use AI tools to attack these questions, but without much notable success\. One potential way forward is to find further proofs of Sendov’s conjecture that utilize other techniques that might be more broadly applicable to this larger family of problems\. The proof here is remarkable in that the zeroes and critical points are treated almost as independent mathematical objects, communicating with each other only very narrowly through four identities in which one only inspects the underlying polynomial \(and its derivative\) at a small number of points\. It could be that an approach focusing on more global features of the polynomial may lead to new proofs of Sendov’s conjecture, and perhaps also of its generalizations\.