Find all polynomials $P$ with real coefficients such that  
$$ \frac{P(x)}{y z}+\frac{P(y)}{z x}+\frac{P(z)}{x y}=P(x-y)+P(y-z)+P(z-x) $$  
for all nonzero real numbers $x, y, z$ obeying $2 x y z=x+y+z$.
The given can be rewritten as saying that
$$
\begin{aligned}
Q(x, y, z) & :=x P(x)+y P(y)+z P(z) \\
& -x y z(P(x-y)+P(y-z)+P(z-x))
\end{aligned}
$$
is a polynomial vanishing whenever \( x y z \neq 0 \) and \( 2 x y z = x + y + z \), for real numbers \( x, y, z \). 

**Claim**: This means \( Q(x, y, z) \) vanishes also for any complex numbers \( x, y, z \) obeying \( 2 x y z = x + y + z \).

Define
$$
R(x, y) := Q\left(x, y, \frac{x+y}{2 x y-1}\right)
$$
which vanishes for any real numbers \( x \) and \( y \) such that \( x y \neq \frac{1}{2} \), \( x \neq 0 \), \( y \neq 0 \), and \( x + y \neq 0 \). This can only occur if \( R \) is identically zero as a rational function with real coefficients. If we then regard \( R \) as having complex coefficients, the conclusion then follows.

**Remark (Algebraic geometry digression on real dimension)**: Note here we use in an essential way that \( z \) can be solved for in terms of \( x \) and \( y \). If \( s(x, y, z) = 2 x y z - (x + y + z) \) is replaced with some general condition, the result may become false; e.g., we would certainly not expect the result to hold when \( s(x, y, z) = x^2 + y^2 + z^2 - (x y + y z + z x) \) since for real numbers \( s = 0 \) only when \( x = y = z \)!

The general condition we need here is that \( s(x, y, z) = 0 \) should have "real dimension two". Here is a proof using this language, in our situation.

Let \( M \subset \mathbb{R}^3 \) be the surface \( s = 0 \). We first contend \( M \) is a two-dimensional manifold. Indeed, the gradient \( \nabla s = \langle 2 y z - 1, 2 z x - 1, 2 x y - 1 \rangle \) vanishes only at the points \( \left( \pm \frac{1}{\sqrt{2}}, \pm \frac{1}{\sqrt{2}}, \pm \frac{1}{\sqrt{2}} \right) \) where the \( \pm \) signs are all taken to be the same. These points do not lie on \( M \), so the result follows by the regular value theorem. In particular, the topological closure of points on \( M \) with \( x y z \neq 0 \) is all of \( M \) itself; so \( Q \) vanishes on all of \( M \).

If we now identify \( M \) with the semi-algebraic set consisting of maximal ideals \( (x - a, y - b, z - c) \) in \( \operatorname{Spec} \mathbb{R}[x, y, z] \) satisfying \( 2 a b c = a + b + c \), then we have real dimension two, and thus the Zariski closure of \( M \) is a two-dimensional closed subset of \( \operatorname{Spec} \mathbb{R}[x, y, z] \). Thus it must be \( Z = \mathcal{V}(2 x y z - (x + y + z)) \), since this \( Z \) is an irreducible two-dimensional closed subset (say, by Krull's principal ideal theorem) containing \( M \). Now \( Q \) is a global section vanishing on all of \( Z \), therefore \( Q \) is contained in the (radical, principal) ideal \( (2 x y z - (x + y + z)) \) as needed. So it is actually divisible by \( 2 x y z - (x + y + z) \) as desired.

Now we regard \( P \) and \( Q \) as complex polynomials instead. First, note that substituting \( (x, y, z) = (t, -t, 0) \) implies \( P \) is even. We then substitute
$$
(x, y, z) = \left(x, \frac{i}{\sqrt{2}}, \frac{-i}{\sqrt{2}}\right)
$$
to get
$$
\begin{aligned}
& x P(x) + \frac{i}{\sqrt{2}} \left( P\left( \frac{i}{\sqrt{2}} \right) - P\left( \frac{-i}{\sqrt{2}} \right) \right) \\
= & \frac{1}{2} x \left( P\left( x - \frac{i}{\sqrt{2}} \right) + P\left( x + \frac{i}{\sqrt{2}} \right) + P(\sqrt{2} i) \right)
\end{aligned}
$$
which in particular implies that
$$
P\left( x + \frac{i}{\sqrt{2}} \right) + P\left( x - \frac{i}{\sqrt{2}} \right) - 2 P(x) \equiv P(\sqrt{2} i)
$$
identically in \( x \). The left-hand side is a second-order finite difference in \( x \) (up to scaling the argument), and the right-hand side is constant, so this implies \( \operatorname{deg} P \leq 2 \).

Since \( P \) is even and \( \operatorname{deg} P \leq 2 \), we must have \( P(x) = c x^2 + d \) for some real numbers \( c \) and \( d \). A quick check now gives the answer \( P(x) = c (x^2 + 3) \) which all work.