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)?

Let $p(t) = t^3 - at^2 + bt - c$ denote the monic polynomial in $\mathbb{R}[x]$ with roots $x$, $y$, $z$. We are given that $$1 = x^2 + y^2 + z^2 = (x+y+z)^2 - 2(xy+yz+xz) = a^2 -2b$$ and asked to show $$ca = xyz(x+y+z) \le 1/3$$ 
By the AM-GM inequality, we know $$\frac{x+y+z}{3} \ge \sqrt[3]{xyz} \implies \frac{a^3}{27} \ge xyz = c$$ so the desired inequality follows if we can prove $a^4 \le 9 \iff a^2 \le 3$.
Since $p(t)$ has three real roots, the derivative $p'(t)$ must have two (with multiplicity) real roots, hence the discriminant of $p'(t)$ is nonnegative:$$p'(t) = 3 t^2 + 2at+b \implies 4a^2 - 12b \ge 0 \implies a^2 \ge 3b$$ Using the constraint $a^2 - 2b = 1$, we have $$a^2 \ge 3 \cdot \frac{a^2 - 1}{2} \implies a^2 \le 3$$ as desired.
Remark: We've seemingly avoided using C-S. However, one way to prove C-S in general is by appealing to a discriminant bound like the above argument. So, we probably haven't avoided C-S as much as covered up our usage with more elementary language.
