<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] I found an alternative proof using the GM-AM inequality and a small observation.
Firstly, using the AM-GM inequality for x,y,z we get  
$x^2yz + xy^2z + xyz^2 = (x+y+z)xyz = (x+y+z)\left(\sqrt[3]{xyz}\right)^3 \leq \frac{(x+y+z)^4}{27} = \frac{1}{3}\frac{(x+y+z)^4}{9}$
Now, notice that: $(x+y+z)^4 \leq 9(x^2 + y^2 + z^2)^2$. This is because we can square-root both sides to get:
$(x+y+z)^2 \leq 3(x^2 + y^2 + z^2) \Leftrightarrow 0 \leq (x-y)^2 + (y-z)^2 + (z-x)^2$
Hence, continuing the first line of the proof we get
$\frac{1}{3}\frac{(x+y+z)^4}{9} \leq \frac{1}{3}(x^2 + y^2 + z^2)^2 = \frac{1}{3}$
and we are done!
Edit: This line of reasoning nicely shows that the equality holds iff $x=y=z$.
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