Show that for all $a,b,c>0$, $\frac 1 {\sqrt[3]{(a+b)(b+c)(c+a)}}\geq\frac 3 {2(a+b+c)}$.

Question

Show that for all $a,b,c>0$, $\displaystyle\frac 1 {\sqrt[3]{(a+b)(b+c)(c+a)}}\geq\frac 3 {2(a+b+c)}$.

I tried to cube the both sides, and expand it, but that'll be too troublesome, is there simpler ways? Thasnk you.

Answer 1

HINT:

Using AM-GM inequality,

$$\frac{a+b+b+c+c+a}3\ge \sqrt[3]{(a+b)(b+c)(c+a)}$$