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.)

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}.
