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

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

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 $$\frac{x+y+z}{3}\ge \sqrt{xyz}$$. 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.

• If you can prove that $HM ≤ GM$, you are done (HM is the harmonic mean). Oct 23, 2020 at 1:35
• Try the same problem with a $3$ instead of $2$ (a bit stricter) and think about AM-GM. Oct 23, 2020 at 1:35

We need to prove that $$xy+xz+yz\geq2\sqrt{x^2y^2z^2},$$ which follows from AM-GM: $$xy+xz+yz\geq3\sqrt{x^2y^2z^2}>2\sqrt{x^2y^2z^2}.$$
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: $$\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}\ge 3\sqrt{\frac{ab.bc.ca}{a^2b^2c^2}}=3>2$$