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.
 A: Lagrange multiplier is the correct tool for solving this type of constrained optimization problem. I don't know your backgroud, but the following is a completely elementary approach (which I only found out with the help of Lagrange).
First it's not hard to show that the maximum (which exists as the sphere is compact) is positive. If the maximum is achieved when two of the three factors are negative, say $x^2-yz<0$ and $y^2-xz<0$, then $yz>x^2\ge 0, xz>y^2\ge 0$, therefore $x,y,z$ all have the same sign. But then we may flip the sign of $z$, so that $(x^2+yz)(y^2+xz)(z^2-xy)\ge (x^2-yz)(y^2-xz)(z^2-xy)$. Therefore the max can only be achieved when all three factors are positive.
Use the geometric mean $\le$ the arithmetic mean, we have $f(x,y,z)^{1/3}\le\frac{x^2+y^2+z^2-xy-yz-xz}{3}=\frac{1-xy-yz-xz}{3}$. And equality holds only when $x^2-yz=y^2-xz=z^2-xy \Leftrightarrow (x-y)(x+y+z)=0, (y-z)(x+y+z)=0$. As $x,y,z$ cannot be all equal, we must have $x+y+z=0$. If we can show further that $\frac{1-xy-yz-xz}{3}$ achieves a maximum when $x+y+z=0$, then we can conclude maximum can be achieved for $f(x,y,z)$ for the same triple.
We are left to find when $\frac{xy+yz+xz}{3}$ achieves minimum on the sphere. This is equivalent to find the minimum of $(x+y+z)^2=(x^2+y^2+z^2) + 2(xy+yz+xz)=1+2(xy+yz+xz)$ on the sphere, and clearly the minimum of  $(x+y+z)^2$ is achieved when $x+y+z=0$.
Put all the work together and reverse the reasoning, we conclude that $f(x)$ achieves a maximum at $(x,y,z)\in \mathbb S^2$ if and only if $x^2-yz, y^2-xz, z^2-xy$ are all positive and $x+y+z=0$. (In fact, e.g. $x^2-yz = y^2+z^2+yz\ge 0$ on $x+y+z=0$, hence the condition of the three factors being positive is redundant. And the intersection of the sphere with the plane are exactly all the solutions to the optimization problem.)
Now we may use any point on the intersection of the plane and the sphere to calculate the maximum. E.g. $x=\frac{1}{\sqrt 2}, y = -\frac{1}{\sqrt 2}, z = 0$, then the maximum is just $\frac{1}{8}$ as pointed out in the comment.
A: 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.
