McLaurin inequality

Question

Let $x,y,z$ be positive real numbers such that $x+y+z=1$ and $xy+yz+xz=\frac{1}{3}$.

What is the value of $\frac{4x}{y+1}+\frac{16y}{z+1}+\frac{64z}{x+1}$?

Asked on Quora

Answer

We can directly compute the values of x, y, and z from the first two equations using Maclaurin’s inequality. It is a refinement of the arithmetic-geometric mean inequality and says that for positive x, y, z we have

$\frac{x+y+z}{3}\ge \sqrt{\frac{xy+yz+xz}{3}}\ge \sqrt[3]{xyz}$

with equality only if $x=y=z$. But since

$\frac{1}{3}=\frac{x+y+z}{3}\ge \sqrt{\frac{xy+yz+xz}{3}}=\sqrt{\frac{1}{9}}=\frac{1}{3},$

we must have equality giving $x=y=z=\frac{1}{3}$. Now we only have to plug in the values to obtain

$\frac{4x}{y+1}+\frac{16y}{z+1}+\frac{64z}{x+1}=\frac{\frac{84}{3}}{\frac{4}{3}}=21$

Here, Terence Tao published a blog post (and research paper) related to Maclaurin’s inequality.