Mathematical Reflections 2017, Issue 4 - Problem J415

Problem:
Prove that for all real numbers $x,y,z$ at least one of the numbers
$$2^{3x-y}+2^{3x-z}-2^{y+z+1}$$ $$2^{3y-z}+2^{3y-x}-2^{z+x+1}$$ $$2^{3z-x}+2^{3z-y}-2^{x+y+1}$$
is nonnegative.

Proposed by Adrian Andreescu, Dallas, Texas, USA

Solution:
Let $a=2^x$, $b=2^y$, $c=2^z$. Adding the three given numbers, we get
$$\begin{array}{lll} S&=&\left(\dfrac{a^3}{b}+\dfrac{a^3}{c}-2bc\right)+\left(\dfrac{b^3}{c}+\dfrac{b^3}{a}-2ca\right)+\left(\dfrac{c^3}{a}+\dfrac{c^3}{b}-2ab\right)\\&=&\left(\dfrac{a^3}{b}+\dfrac{b^3}{a}-2ab\right)+\left(\dfrac{b^3}{c}+\dfrac{c^3}{b}-2bc\right)+\left(\dfrac{c^3}{a}+\dfrac{a^3}{c}-2ca\right). \end{array}$$
By the AM-GM Inequality, we get
$$\dfrac{a^3}{b}+\dfrac{b^3}{a} \geq 2ab, \qquad \dfrac{b^3}{c}+\dfrac{c^3}{b} \geq 2bc, \qquad \dfrac{c^3}{a}+\dfrac{a^3}{c} \geq 2ca.$$
So, $S \geq 0$ and we conclude that at least one of the three given numbers is nonnegative.