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

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!
