British Maths Olympiad (BMO) 2002 Round 1 Question 3 Proof without Cauchy-Schwarz?

Yes, we can can get a sum of squares here. We need to prove that $$(x^2+y^2+z^2)^2\geq3xyz(x+y+z)$$ or $$\sum_{cyc}(x^4+2x^2y^2-3x^2yz)\geq0$$ or $$\sum_{cyc}(2x^4-2x^2y^2+6x^2y^2-6x^2yz)\geq0$$ or $$\sum_{cyc}(x^4-2x^2y^2+y^4+3(x^2z^2-2z^2xy+y^2z^2))\geq0$$ or $$\sum_{cyc}((x^2-y^2)^2+3z^2(x-y)^2)\geq0.$$

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 $\,x=y=z\,$.

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.