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)?
 A: 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
$$\sum_{cyc}((x^2-y^2)^2+3z^2(x-y)^2)\geq0.$$
A: 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 $\,x=y=z\,$.
A: 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.
A: 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 $x=y=z$.
A: 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.
