This post concerns the following conjecture of Sendov, as well as its strengthening by Phelps–Rodriguez:
Conjecture 1 (Sendov’s conjecture) Let
, and let
be a degree
polynomial with all zeroes in the unit disk. Then for every zero
of
, there exists a critical point
of
with
.
Conjecture 2 (Phelps–Rodriguez conjecture) Let
, and let
be a degree
polynomial with all zeroes in the unit disk. Then for every zero
of
, there exists a critical point
of
with
, unless
is on the unit circle and
is a scalar multiple of
.
By applying a rotation around the origin, we can normalize
to be a real number with
.
From the work of Rubinstein, both conjectures were already established in the
case, so one can restrict to the
case. Both of these conjectures then follow from
Conjecture 3 (Sendov’s conjecture in interior) Let
. Let
be a degree
polynomial with all zeroes in the unit disk. Then if
is a zero of
, there exists a critical point
of
with
.
All three of these conjectures were established for
(in a sequence of papers culminating in this paper of Brown and Xiang) and for sufficiently large
(in a paper of myself, which in turn built upon several partial results in this setting). This left the case of intermediate
to be settled. My arguments used some qualitative ingredients (most notably analytic continuation) and as such did not easily lend themselves to quantifying the threshold of
above which the argument was valid.
Recently, Lech Mazur was able to use an AI tool to resolve Sendov’s conjecture for all
, with the proof verified in Lean. 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) 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, 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 (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 the entire argument in Lean, extended to
by 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. The
cases 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 degree
polynomial
with zeroes

for some

and

,

in the closed unit disk, whose critical points all lie a distance at least

from

. We use

notation here in the non-asymptotic sense, thus

means that

for some absolute constant

(independent of

). We will also use the notation

to denote a quantity that is bounded in magnitude by

.
To capture the fact that the critical points lie at a distance at least
from
, we write these critical points as

for some (non-zero)

in the closed unit disk.
Example 4 If
and
, then
are the non-trivial
roots of unity, while the
are all equal to
. Strictly speaking this is not actually a counterexample to Conjecture 3, because
is 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. Here one took 
where
was an asymptotic parameter going to infinity, 
was a low-degree polynomial for some
, 
and
were constants. This polynomial has a zero at
,
critical points at
, and
additional 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 that 
Here
denotes a quantity that goes to zero as
. If one ignores the
errors, one can show that these conditions are only simultaneously feasible if
and all the
vanish, but the argument was somewhat subtle (I had to proceed by inspecting the second Fourier coefficient of (1)). This illustrates the fact that the regime
is particularly delicate.
We now have two sets of points in the closed unit disk:
and
. They “communicate” with each other through the polynomial
and its first derivative
, both of which can be expressed in terms of either set of points (as well as
and
). Indeed, if we normalize
to be monic, then we can factor
in 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 both

and

, 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 case
right away. In this case we see from (3), (4) 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 inspecting
or
at various natural locations, we can thus obtain a number of identities relating the
to the
. We record the ones that we actually need here:
Lemma 6 (Communication identities) Let
denote the function 
- (i) (Centroid identity) We have

That 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 the
vanish.)
Proof: For (i), we inspect the behavior of
as
. From (2) we have

and thus on differentiating term by term

Meanwhile, from
(4) we have

Comparing coefficients, we obtain the claim.
For (ii), we consider the expression
. On the one hand, from (2), (3) one has

(Note from hypothesis that

cannot be a critical point, so the denominator is non-zero.) On the other hand, from
(4),
(5) one has

Equating the two identities, we obtain (ii) after some algebra.
For (iii), we evaluate
. From (2) we have

while from
(5) we have

Equating the two identities, we obtain (iii) after some algebra using
(6).
For (iv), we similarly evaluate
. From (3) we have

while from
(4) one has

Equating the two identities, we obtain (iv) after some algebra using
(6).
Remarkably, the polynomial
will play no further role in the argument: the identities in (i)-(iv), together with the hypotheses that
and
lie in the closed unit disk, will be sufficient by themselves to obtain a contradiction.
Example 7 Continuing the example in Example 4, 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. The comparison of the polynomial at a location
and at the polar inversion
of that location across the closed unit disk is a familiar trick in the literature; see, e.g., Lemma 5 and Theorem 8 of Dégot. The specific form of the polar identity is implicit in the first part of Section 5 of Mazur’s AI-generated proof, 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 and (in the
case) Rubinstein. The work of Meir–Sharma and Mir–Nazir–Wani also contain several further identities relating the
to the
; see in particular Lemma 15 below. Variants of (5) also appear in Proposition 10 of Miller.
Remark 9 The first origin identity (9) is already strong enough to handle asymptotically all examples of the form in Example 5, except in the endpoint case where
vanish and the
are all
. Indeed, as the
are in the closed unit disk, (9) 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 the
all vanish.
As mentioned in Example 5, the most delicate regime occurs when
. It is convenient to introduce the normalized version

of

, thus

, and the case

corresponds to

. Informally,

measures how close

is to

(at the scale of

).
A key role in the argument will be played by the mean

of the

, particularly the real part

. As the

all lie in the unit disk, the mean

does also, so that

and

On the other hand, in the example in Example
4,

is equal to the extremal value of

, and

. In Example
5, 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 of

as a measure of how close

is to

. Clearly we have

for all

(note that

is strictly less than

).
The arguments will revolve around the relationship between
and
. 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)
It will be the inequality (18) that we use in practice, but it will be derived from (17), which in turn is a consequence of (16), which will follow from the polar identity (8) together with the fact that the
and
lie in the unit disk. The bound (18) is only slightly weaker than (17); 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 contains several similar inequalities. The inequality (16) was extracted from (5.1) of Mazur’s AI-generated proof; the subsequent bounds (17), (18) 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
.
Part (i) (which was extracted with some effort from Section 6 of the original AI-generated argument) will be deduced from the first and second origin identities (9), (10), as well as the centroid identity (7). Part (ii) will follow from (i) and the polar inequality (18), while part (iii) is an elementary consequence of (i).
As it turns out, the last three terms in (21) are asymptotically negligible as
. Dropping those terms gives a competing feasibility region for
and
which is disjoint from the one coming from the polar inequality (17) (or (18)):
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) added, there is enough room between the two inequalities (18), (21) to obtain a contradiction for all
(using the additional bound
to 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) is monotone increasing in
(or equivalently, monotone decreasing in
). In view of (18), we can thus replace
by
in this inequality, so that
is replaced by

and

replaced by

. The inequality
(21) then becomes an inequality involving only

and

:

We also note that the bounds

force the constraint

This prevents

from getting too close to the upper limit

(or

getting too close to zero).
We can now eliminate all large degrees, e.g.,
, as follows. The quadratic
attains its minimum at
. For
we have

while for

(if this region is non-vacuous) we can bound the quadratic by its value

at

. Thus

Evaluating these expressions, we arrive at

Since

, we have

. Next, we claim that

. As

is monotone increasing in

, it suffices to do this when

. Here one can directly compute that

since the discriminant

of the numerator is negative, we conclude that

as desired.
Dropping some
and
terms, we conclude that

Every term on the right-hand side can be seen to be decreasing in

for

. 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) remains infeasible in this range. This can be illustrated numerically without much difficulty: see
this applet. For instance, in the most delicate case

, the right-hand side of
(22) only gets as large as

(and in particular stays below

) throughout the range

:
I have also verified this bound in Lean.
— 1. The polar inequality —
We begin with a proof of Proposition 10.
As is well known, the Möbius transform
maps the closed unit disk to itself. In particular, we have

for all of the zeroes

. Inserting this into the polar identity
(8) and using the triangle inequality, we conclude the lower bound

We now convert this bound to a bound involving the quantity

in
(12). From the arithmetic mean-geometric mean inequality we have

and from
(12) we have

Since

, we thus have

giving the raw polar inequality
(16).
Bounding
by
and using the quantities
from (11), (15), we observe that

Using the basic inequality

, we thus have

with strict inequality for

. From
(16) we conclude
(17). This also implies

, since otherwise the integrand is always bounded by

, which is absurd.
On evaluating the integral in (17), 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

where

and

. If we can show that

for all

, then taking logarithms in
(17) yields

from which
(26) will follow by routine algebra.
Both sides of (27) vanish at
. Taking derivatives, it suffices to show that

which rearranges to

To expand the left-hand side, we use the double angle formulae

and

to rewrite it as

Collecting the coefficient of

for

and 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) is only slightly weaker than (17). For small
, one can perform Taylor approximation on the latter bound to obtain 
while the former bound is 
Note that
is slightly smaller than
.
Relating to this, the constant
in (27) cannot be improved.
— 2. The origin inequality —
Now we turn to the proof of Proposition 11, which is more difficult and revolves around an analysis of the function
defined in (6). We begin with a heuristic analysis. Inserting the approximation
for small
into (6) and using (12), we are led to the approximation

at least when

is small (which turns out to be the dominant regime in applications). This suggests a relation

between the two expressions involving

in the origin identities in Lemma
6. Substituting in this approximation, we obtain some (slightly complicated) approximation for the sum

in terms of

,

,

,

, and the product

.
As
lie in the closed unit disk, the product
does also. However, past experience with the Sendov conjecture has taught us that the worst cases tend to be when
lie very close to the boundary of the disk, so that
is close to one. For instance, in Example 4 all the
and
lie on the unit circle, and
. See Remark 3 of Dégot or Theorem 1.10(ii) of my own paper for other places where this heuristic is noted. To simplify the discussion, let us assume for now that
is exactly one, so that
all lie on the unit circle. This leads in particular to the inversion identities

The centroid identity in Lemma
6(i) relates the sum of the

with the sum of the

. Using
(31), this gives a similar identity relating the sum of the

with 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 of

and

are minor and can be quickly eliminated for the purposes of obtaining useful bounds, leading eventually to the relation in Proposition
11.
We turn to the details. To make the approximation (30) more precise, we note that
, and hence by the fundamental theorem of calculus

The heuristic
(29) predicts that

, which would give
(30). If we actually differentiate
(6) carefully, we obtain the exact identity

Bounding

, we write this

When faced with a similar expression in (24), we used the arithmetic mean-geometric mean inequality. Here, the analogous tool is Maclaurin’s inequality, which gives

and hence by Cauchy–Schwarz

Repeating the calculations used to show
(25), we have

and so we obtain the bound

Integrating this, we obtain a rigorous analogue of
(30),

and thus by the triangle inequality

From the first and second origin identities
(9),
(10) we have

The next step is thus to estimate
. When
, then all the
were on the unit circle and we could use (31) (and the centroid identity) to proceed. Now, we are no longer assuming
to equal
, but we can still adapt the previous arguments with a loss proportional to
. The key lemma is
Lemma 13 (Defect lemma) Let
be some points in the closed unit disk. Then 
Proof: By a limiting argument we may assume that none of the
vanish. If we write
for some
, then we can calculate that

and

Thus the desired inequality reduces to the superadditivity property

But from the sinh addition formula

we have

for all non-negative

(this also follows from the convex nature of

together 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) 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 the

vanish. Applying
(12) and some algebraic manipulation, we arrive at

Substituting this back into
(33), we conclude that

and hence after some algebra and the triangle inequality

Inserting this into
(32), we obtain

We can simplify
(34) by reducing to the

case. Indeed, we shall show that

which implies that the right-hand side of
(34) is non-decreasing in

in the range

. Thus we may replace

by

in
(34) to conclude that

Let us now verify
(35). Using

and the triangle inequality, we can lower bound

Inserting this into
(35) 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 fact

suffices), giving the claim
(35).
Next we eliminate the role of the imaginary term
. Observe for any complex number
with positive real part that

as can be seen by squaring both sides. The expression

has real part

which lies between

and

(in particular, it is positive), and imaginary part of magnitude at most

by
(13). We conclude that

The right-hand side can be rearranged using the quantity

from
(11) as

so the bound
(36) gives
(19).
— 2.1. Upper bound on
—
Now we can prove (20). Suppose for contradiction that
; since
, this implies that
. Crudely discarding the
term in (19) and bounding
by
, we have

The quadratic polynomial

equals

at

and attains its minimum at

with value

. By convexity, we thus have

for

and

for

(this latter statement is vacuous if

). Since

, we can therefore crudely bound


and hence

From
(15) we have

, thus by
(18) one has

From another application of
(18) one has

We conclude that

It is now convenient to introduce the quantity

, thus

with

and

Inserting these bounds and dividing by

, we conclude

Since

, we obtain

Since

and

, we conclude that

Routine calculus shows that

has a maximum of at most

, and that the right-hand side here is at most

, giving the required contradiction. This proves
(20).
— 2.2. A simplified estimate —
Now we show (21). Note from (11) that

while from
(15) we have

and hence also

From
(15) we have

By the mean value theorem (noting that

is non-negative) we thus have

From the standard beta function identity

(and the fact that

) we can thus replace
(19) by

From
(15) we have

Thus by
(37),
(38),
(39) 
Dividing by the positive quantity

gives 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 the
case, i.e.,
Theorem 14 (Rubinstein’s theorem) Let
, and let
be a degree
polynomial with all zeroes in the unit disk. If
, then there exists a critical point
of
with
, unless
is a scalar multiple of
.
The argument here is essentially in Remark 5.1 of this paper of Tang and Zhang.
Taking contrapositives, we may assume that the critical points
are of the form
for some
in the closed unit disk, and normalize
to be monic; our task is to show that
.
The polar identity (8), based on calculating
degenerates 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:
Lemma 15 (Meir–Sharma identity) If
and the critical points are of the form
then all the zeroes
are not equal to
, and 
Proof: By hypothesis,
is not a critical point of
, so
and
for all
. Instead of computing
, we instead consider the expression
. On the one hand, from (4) we have

while from differentiating
(4) we have

Meanwhile, from
(3) we have

and from differentiating
(3) we have

Using these identities to compute

in two different ways gives the claim.
Now take
. Since
lie in the closed unit disk,
has real part at least
, while
is at most
. Thus, the only way that the above identity can hold is if
for all
, hence
for all
. Thus all critical points are at the origin, which forces
for some
. Since
, we conclude that
, giving the claim.
— 4. The
cases —
We now prove the
cases of Conjecture 3. The starting point is (23). Using the triangle inequality and
, this implies that

(This also follows from
(16) and

.) From Hölder’s inequality and

we conclude that

The right-hand side can be computed to equal

which is obviously less than

for

, 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) Let
and
, and let
be a degree
polynomial with zeroes
satisfying
. Then for every zero
of
, there exists a critical point
of
with
.
Sendov’s conjecture is the limiting case
of this conjecture. There has been relatively little progress on this conjecture: the cases
were established by Khavinson, Pereira, Putinar, Saff, and Shimorin, and in this previous paper 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 the
are 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:
Conjecture 18 (Schmeisser’s conjecture) Let
, and let
be a degree
polynomial with all zeroes in the closed unit disk. Then for any
in the convex hull of the zeroes of
, there exists a critical point
of
with
.
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, the
are 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. A slightly different strengthening was also proposed in Conjecture 1.10 of Tang and Zhang:
Conjecture 19 (Tang–Zhang conjecture) Let
, and let
be a degree
polynomial 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, the
case is the strongest form of the conjecture.
Another well known variant of Sendov’s conjecture is Smale’s problem:
Conjecture 20 (Smale’s problem) Let
, and let
be a degree
polynomial. Then for any zero
of
, there exists a critical point
of
with
.
The constant
is best possible, as can be seen by the example
and
. Using the Koebe one-quarter theorem, Smale proved this conjecture with
replaced by
. Some slight improvements of this bound have been obtained over the years; for instance for
, the improved bound of
was obtained by Crane. 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.