Tricky Inequality

Algebra Level 3

Let $a$ , $b$ , and $c$ be positive real numbers such that $abc=1$ . Find the minimum value of:

$K=\frac{1}{a(1+b)} + \frac{1}{b(1+c)}+ \frac{1}{c(1+a)}$

1 solution

J Dim
Apr 4, 2018

Let $a=\frac{x}{y}$ , $b=\frac{y}{z}$ , $c=\frac{z}{x}$ , $(x,y,z >0)$

(We can do that, since $abc=1$ )

Then we replace on $K$ and after simplifying we get:

$K= \frac{yz}{xy+xz}$ + $\frac{zx}{xy+yz}$ + $\frac{xy}{yz+zx}$

By Nesbitt's well-known inequality for the triple $(xy,yz,zx)$ the minimum value of $K$ is $\frac{3}{2}$ =1.5