Question

Prove $\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\ge a+b+c,$ for $a,b,c>0$

Answer 1

Using $AM\ge GM$, for positive quantities,

$\frac{\frac{bc}{a}+\frac{ca}{b}}{2}\ge \sqrt{\frac{bc}{a}\cdot \frac{ca}{b}}\Rightarrow \frac{bc}{a}+\frac{ca}{b}\ge 2c\left(1\right)\frac{\frac{ca}{b}+\frac{ab}{c}}{2}\ge \sqrt{\frac{ca}{b}\cdot \frac{ab}{c}}\Rightarrow \frac{ca}{b}+\frac{ab}{c}\ge 2a\left(2\right)\frac{\frac{bc}{a}+\frac{ab}{c}}{2}\ge \sqrt{\frac{bc}{a}\cdot \frac{ab}{c}}\Rightarrow \frac{bc}{a}+\frac{ab}{c}\ge 2b\left(3\right)$

Adding $\left(1\right),\left(2\right),\left(3\right)$

$2[\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}]\ge 2(a+b+c)[\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}]\ge (a+b+c)$