Prove that $(x-y)(y-z)(z-x) \leq \frac{1}{\sqrt{2}}$

Question

If $x,y,z$ are real and $x^2+y^2+z^2=1$, prove that$$(x-y)(y-z)(z-x) \leq \frac{1}{\sqrt{2}}.$$

Equality is achieved in some strange cases: For example, if $x = -\dfrac{1}{\sqrt{2}}$, $y = 0$ and $z = \dfrac{1}{\sqrt{2}}$, then $(x-y)(y-z)(z-x)=\dfrac{1}{\sqrt{2}}$.

Note that the claim is obvious in the case when $x \geq y \geq z$ (since $(x-y)(y-z)(z-x)\leq 0$ in this case). But the inequality is not symmetric in $x, y, z$ (only cyclic). Thus, you cannot WLOG assume that $x \geq y \geq z$. (But you can WLOG assume that $x \leq y \leq z$ because of the previous observation.)

Answer 1

If $\prod\limits_{cyc}(x-y)<0$ then it's obvious.

But for $\prod\limits_{cyc}(x-y)\geq0$ it's enough to prove that $$(x^2+y^2+z^2)^3\geq2(x-y)^2(x-z)^2(y-z)^2.$$

Now, let $x\leq y\leq z$, $y=x+u$ and $z=x+v$.

Thus, we need to prove that $$(3x^2+2(u+v)x+u^2+v^2)^3\geq2(u-v)^2u^2v^2$$ or $$3x^2+2(u+v)x+u^2+v^2-\sqrt[3]{2(u-v)^2u^2v^2}\geq0,$$ for which it's enough to prove that $$(u+v)^2-3\left(u^2+v^2-\sqrt[3]{2(u-v)^2u^2v^2}\right)\leq0$$ or $$2(u^2-uv+v^2)\geq3\sqrt[3]{2(u-v)^2u^2v^2},$$ which is true by AM-GM: $$2(u^2-uv+v^2)=2(u-v)^2+uv+uv\geq3\sqrt[3]{2(u-v)^2u^2v^2}.$$ Done!

Answer 2

Here is a proof similar to the solution of IMO 2006 problem 3 (well... at least to the solution I found on the IMO):

As Michael Rozenberg noticed, it suffices to prove the inequality \begin{equation} \left( x^2 +y^2 +z^2 \right) ^3 \geq2\left( x-y\right) ^2 \left( y-z\right) ^2 \left( z-x\right) ^2 . \label{darij1.eq.goal1} \tag{1} \end{equation} Indeed, if this inequality is proved, then we can take square roots on both sides, and obtain \begin{align*} \sqrt{\left( x^2 +y^2 +z^2 \right) ^3 } & \geq\sqrt{2\left( x-y\right) ^2 \left( y-z\right) ^2 \left( z-x\right) ^2 }\\ & =\left\vert \sqrt{2}\left( x-y\right) \left( y-z\right) \left( z-x\right) \right\vert \geq\sqrt{2}\left( x-y\right) \left( y-z\right) \left( z-x\right) , \end{align*} which is the claim of the original post in a homogenized form.

Here is the idea of the proof of \eqref{darij1.eq.goal1} (see below for the implementation): We notice that if we replace $x,y,z$ by $x+p,y+p,z+p$ for a fixed $p\in\mathbb{R}$, then the right hand side of \eqref{darij1.eq.goal1} does not change, but the left hand side may become smaller if we pick $p$ appropriate. The trick is to pick $p$ such that the left hand side becomes as small as possible. This will make the inequality \eqref{darij1.eq.goal1} sharper and easier to prove (spoiler: the two sides will differ by a single square).

Picking the right $p$ is easy: The sum $\left( x+p\right) ^2 +\left( y+p\right) ^2 +\left( z+p\right) ^2 $ is a quadratic polynomial in $p$, and is easily seen to be minimized for $p=-\dfrac{x+y+z}{3}$.

Let us actually do this. So let us set $p=-\dfrac{x+y+z}{3}$. Then, \begin{equation} x^2 +y^2 +z^2 \geq\left( x+p\right) ^2 +\left( y+p\right) ^2 +\left( z+p\right) ^2 , \label{darij1.eq.side1} \tag{2} \end{equation} because a straightforward computation shows that \begin{align*} & \left( x^2 +y^2 +z^2 \right) -\left( \left( x+p\right) ^2 +\left( y+p\right) ^2 +\left( z+p\right) ^2 \right) \\ & =\dfrac{1}{3}\left( x+y+z\right) ^2 \geq0\qquad\left( \text{since squares are nonnegative}\right) . \end{align*} Both sides of the inequality \eqref{darij1.eq.side1} are nonnegative reals; thus, we can take them to the $3$-rd power and obtain \begin{equation} \left( x^2 +y^2 +z^2 \right) ^3 \geq\left( \left( x+p\right) ^2 +\left( y+p\right) ^2 +\left( z+p\right) ^2 \right) ^3 . \label{darij1.eq.side13} \tag{3} \end{equation} But a straightforward (if ugly) computation shows that \begin{align*} & \left( \left( x+p\right) ^2 +\left( y+p\right) ^2 +\left( z+p\right) ^2 \right) ^3 -2\left( x-y\right) ^2 \left( y-z\right) ^2 \left( z-x\right) ^2 \\ & =\dfrac{2}{27}\left( 2x-y-z\right) ^2 \left( 2y-z-x\right) ^2 \left( 2z-x-y\right) ^2 \geq0 \end{align*} (since squares are nonnegative). Hence, \begin{equation} \left( \left( x+p\right) ^2 +\left( y+p\right) ^2 +\left( z+p\right) ^2 \right) ^3 \geq2\left( x-y\right) ^2 \left( y-z\right) ^2 \left( z-x\right) ^2 . \end{equation} Combining this with \eqref{darij1.eq.side13}, we find $\left( x^2 +y^2 +z^2 \right) ^3 \geq2\left( x-y\right) ^2 \left( y-z\right) ^2 \left( z-x\right) ^2 $. This proves \eqref{darij1.eq.goal1}.