Prove that if $x, y, z$ are positive then $\frac{2}{1/x+1/y+1/z} \le \sqrt[3]{xyz}$

Question

Prove that if $x, y, z$ are positive then $$\frac{2}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \le \sqrt[3]{xyz}$$

I need to prove that using only basic facts about inequalities. This was supposed to be at the beggining of a calculus course so nothing advanced should be expected. I have previously proved the GM-AM inequality for $n=3$, so I get the feeling this can be solved somewhat using the fact that $\frac{x+y+z}{3}\ge \sqrt[3]{xyz}$. I also thought that if I was able to prove that $8x^2y^2z^2\le(x+y+z)^3$ that would give me a straight forward solution but it seemed like a death end.

I would like some hints, I just can't find a useful property to start with the proof.

Answer 1

We need to prove that $$xy+xz+yz\geq2\sqrt[3]{x^2y^2z^2},$$ which follows from AM-GM: $$xy+xz+yz\geq3\sqrt[3]{x^2y^2z^2}>2\sqrt[3]{x^2y^2z^2}.$$

Answer 2

let $x=a^3,y=b^3,z=c^3$

we have to prove on rearranging $$\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}>2$$

which is true because: $$\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}\ge 3\sqrt[3]{\frac{ab.bc.ca}{a^2b^2c^2}}=3>2$$