Lower bound of a constrained 3-variables expression

Question

Let $x,y,z$ be non-negative reals such that $x+y+z = 1$.
Does this imply $$\begin{align} 0\: & \leqslant\:(x+y-z)(x-y)^2\:+\:(y+z-x)(y-z)^2\:+\:(z+x-y)(z-x)^2 \\[2ex] & =\:\sum_\text{cyc}(1-2z)(x-y)^2 \\[1ex] & =\:\sum_\text{cyc}(x-y)(1-2z)(x-y)\;?\end{align}$$ Negative contributions to the sum may arise if one of the variables exceeds $\frac12$.

The expression of third degree involving three variables points towards Schur's inequality.
If the estimate is true and Schur could be applied, do you see nevertheless a proof which avoids Schur's inequality?

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The intention to avoid Schur is due to the fact that no matrix version of it is known $-$ as far as I know: When generalising to the situation where $x,y,z$ are positive-semidefinite matrices (or positive-semidefinite operators on an infinite-dimensional Hilbert space), then the above sum as given in the last formula line equals a Hermitian matrix. It would be interesting to know its sharp lower bound of the form $\,m\mathbb 1$, and this may require a scalar $m$ which is strictly smaller than zero.

Answer 1

This is the third degree form of Schur's inequality. If you want to avoid it, a known factorization for Schur's inequality is:

$$(x+y-z)(x-y)^2\:+\:(y+z-x)(y-z)^2\:+\:(z+x-y)(z-x)^2=2\left[xyz-(x+y-z)(x-y+z)(-x+y+z)\right]$$

So you need to prove:

$$xyz \geq (x+y-z)(x-y+z)(-x+y+z)$$

WLOG, let $x\geq y\ge z$. Then $x+y-z\geq 0$ and $x-y+z \geq 0$. If $-x+y+z <0$, we have $LHS \geq 0 \geq RHS$. Otherwise, $x, y$ and $z$ are side-lengths of a triangle, and we can apply AM-GM as follows:

$$\sqrt{(x+y-z)(x-y+z)} \leq \frac{(x+y-z)+(x-y+z)}{2}=x$$

And now multiply with the other two similar inequalities to complete the proof.

Answer 2

Let $x\geq y\geq z$.

Thus, $1-2z\geq0$, $1-2y=x+z-y\geq0$ and $$\sum_{cyc}(1-2z)(x-y)^2\geq(x-z)^2(1-2y)+(y-z)^2(1-2x)\geq$$ $$\geq(y-z)^2(1-2y)+(y-z)^2(1-2x)=2(y-z)^2(1-y-x)\geq0.$$