British Maths Olympiad 2002 Round 1 Problem 1

My solution to an Olympiad problem.

By Vamshi Jandhyala in mathematics

September 29, 2019

Let $x, y, z$ be positive real numbers such that $x^{2} + y^{2} + z^{2} = 1$. Prove that $x^{2}yz + xy^{2}z + xyz^{2} \leq \frac{1}{3}$.

Solution

Using Muirhead’s inequality, we have

$$ \begin{align} 2(x^{2}yz + xy^{2}z + xyz^{2}) \leq 2(x^{4}+y^{4}+z^{4}) \label{y2002r1p1e1}\\
2(x^{2}yz + xy^{2}z + xyz^{2}) \leq 2x^{2}y^{2}+2x^{2}z^{2}+2y^{2}z^{2} \label{y2002r1p1e2} \end{align} $$

We have $\ref{y2002r1p1e1}$ because (4,0,0) majorizes (2,1,1).

We have $\ref{y2002r1p1e2}$ because (2,2,0) majorizes (2,1,1).

Adding $\ref{y2002r1p1e1}$ and $\ref{y2002r1p1e2}$, we get

$$ \begin{align} 3(x^{2}yz + xy^{2}z + xyz^{2}) \leq (x^{2}+y^{2}+z^{2})^{2} = 1 \\
\end{align} $$