British Maths Olympiad (BMO) 2002 Round 1 Question 3 Proof without Cauchy-Schwarz? The question states:
Let $x,y,z$ be positive real numbers such that $x^2 + y^2 + z^2 = 1$
Prove that
$x^2yz + xy^2z + xyz^2 ≤ 1/3$
I have a proof of this relying on the fact that:
$x^2/3 +y^2/3 + z^2/3 \geq (x+y+z)^3/9  $ (A corollary of C-S I believe)
Is there an elementary proof without this fact (or C-S in general)?

Here is my approach.
Using AM-GM,
$\begin{align} x^2+y^2+z^2 &\ge 3(xyz)^{2/3} \\ \implies 1 &\ge 3(xyz)^{2/3} \\ \implies (xyz)^{2/3} &\le \dfrac{1}{3} \\ \implies xyz \le \dfrac{1}{3\sqrt{3}} \tag 1 \end{align}$
Again, using AM-GM,
$\begin{align} \dfrac{x^3+y^3+z^3}{3} &\ge 3xyz \\ \implies x^3+y^3+z^3-3xyz &\ge 0 \\ \implies (x+y+z)(x^2+y^2+z^2 -xy-yz-zx) &\ge 0\end{align}\tag*{}$
Since $x,y,z \in \mathbb R^+ $, $x+y+z \ge 0 $.
$\begin{align}\therefore \, x^2+y^2+z^2 -xy-yz-zx &\ge 0 \\ \implies -x^2-y^2-z^2 +xy+yz+zx &\le 0 \\ \implies -2x^2-2y^2-2z^2 +2xy+2yz+2zx &\le 0 \\ \implies x^2+y^2+z^2 +2xy+2yz+2zx &\le 3(x^2+y^2+z^2) \\ \implies (x+y+z)^2 &\le 3 \\ \implies x+y+z &\le \sqrt{3} \tag 2 \end{align}$
Multiplying $(1)$ and $(2)$, we can get the desired result.
