<s>[INST] Finding the maximum value of the following product. 
Let $x,y,z$ be three reals which satisfies $x^2+y^2+z^2=1$. Find the maximum value of $$P = (x^2-yz)(y^2-zx)(z^2-xy).$$

I think expanding would be a bad idea. I tried to apply the Weirstrass product inequality but it may give the minimum not maximum value.
 [/INST] We claim that $P \leq \dfrac{1}{8}.$
Case $1:$ All three terms are greater than or equal to zero.
By AM-GM,
\begin{align}
\sqrt[3]{P} & \leq \dfrac{x^2+y^2+z^2-xy-yz-xz}{3} \\
&= \dfrac{1-(xy+yz+xz)}{3}.
\end{align}
It thus suffices to prove $\dfrac{1-(xy+yz+xz)}{3} \leq \dfrac{1}{2} \iff xy+yz+xz \geq \dfrac{-1}{2}.$
But the last inequality is easy, since
\begin{align}
(x+y+z)^2 \geq 0 & \Rightarrow 1+2(xy+yz+xz) \geq 0 \\
& \Rightarrow xy+yz+xz \geq \dfrac{-1}{2}.
\end{align}
In particular, the maximum value of $P$ is attainable when we have, say, $x=0, y=\dfrac{\sqrt{2}}{2}, z=\dfrac{-\sqrt{2}}{2}.$
Case $2$ : At least one term is smaller than zero.
W.L.O.G. let $z^2-xy <0.$ Now, if all three terms are smaller than zero, then $P < 0 < \dfrac{1}{8}.$ And similarly if only $z^2-xy <0.$ Thus, there must be exactly two terms smaller than zero, and let $y^2-zx <0$ too. Hence, by AM-GM, this gives us
$$\sqrt[3]{P} \leq \dfrac{x^2-yz+zx-y^2+xy-z^2}{3},$$
so it suffices to prove
\begin{align}
\dfrac{x^2-yz+zx-y^2+xy-z^2}{3} \leq \dfrac{1}{2} &\iff x^2-(y^2+z^2)-yz+xz+xy \leq \dfrac{3}{2} \\ 
& \iff x^2-(y+z)^2+xz+xy+yz \leq \dfrac{3}{2} \\
& \Leftarrow (x+y)(x+z) \leq \dfrac{3}{2}
.
\end{align}
Again, by AM-GM, $(x+y)(x+z) \leq \dfrac{(x+y)^2+(x+z)^2}{2}.$ Hence we are done if we could prove
\begin{align}
(x+y)^2+(x+z)^2 \leq 3(x^2+y^2+z^2) & \iff 2xy+2xz \leq x^2+2y^2+2z^2 \\
& \iff x^2+y^2+z^2-2yz-2xz+y^2+z^2 \geq 0 \\
& \iff (x-y-z)^2 -2yz +y^2+z^2 \geq 0 \\
& \iff (x-y-z)^2+(y-z)^2 \geq 0,
\end{align}
Which is obvious.
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