<s>[INST] 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)?
 [/INST] By the RMS-AM-GM inequalities (root-mean square vs. arithmetic vs. geometric means):
$$
\frac{1}{\sqrt{3}} = \sqrt{\frac{x^2+y^2+z^2}{3}} \;\ge\; \frac{x+y+z}{3} \;\ge\; \sqrt[3]{xyz} \quad\implies\quad \begin{cases}\begin{align}x+y+z \,&\le\, \sqrt{3} \\[5px] xyz \,&\le\, \dfrac{1}{3\sqrt{3}}\end{align}\end{cases}
$$
Multiplying the latter gives $\,xyz(x+y+z) \le \dfrac{1}{3}\,$, which is the inequality to prove. As with all means inequalities, the equality holds iff $\,x=y=z\,$.
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