# 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)?

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 $$\sum_{cyc}((x^2-y^2)^2+3z^2(x-y)^2)\geq0.$$

• Nice you were able to avoid use of AM-GM too! Out of curiousity how could you tell that $(x^4+2x^2y^2−3x^2yz)$ would factorise well? – Abe Apr 28 '18 at 20:13
• Not OP but cyclic polynomial sums have a lot more freedom than regular polynomials, since any individual term can have each of it's variables cycled. So basically, if you have a cyclic polynomial sum like this, you might just try to force it into something factorable. Notice that the step of multiplying everything by 2 doubles the number of terms and so doubles the amount of options you have to manipulate the polynomial. This is just one way of taking advantage of the symmetry of the problem! – Alex Jones Apr 29 '18 at 2:30
• General remark: one of the proofs of CS basically rewrites the inequality as a sum of squares $\sum_{i,j} (a_ib_j - a_jb_i) \geq 0$, so every application of CS can be rewritten as a sum of squares in this way. – Federico Poloni Apr 29 '18 at 14:27
• Curious how did you come up with the LHS in the first equation? I wouldn't of thought of it. Is it just from experience/practice? – user557278 Apr 29 '18 at 15:43
• @hojusaram It's just homogenization. It helps sometimes in polynomial inequalities. – Michael Rozenberg Apr 29 '18 at 17:46

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.

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 $x=y=z$.

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.