Proof inequality for all positive numbers How to prove this inequality: it started with
prove:
$\sqrt {ab} \geqslant \frac {2} {1/a+1/b} $  for all positive numbers on $a$ and $b$. Leftside = rightside when $a=b$.
I could make it until:
$\sqrt a \cdot \sqrt b \geqslant \frac {2ab} {b+a} $
this is as far as I could go, I don't know how to prove it further.
 A: You've arrived at $$\sqrt{ab} \geqslant \frac{2ab}{a + b}.$$
The above is equivalent to $$1 \geqslant \frac{2\sqrt{ab}}{a + b}$$ or $$a + b \geqslant 2\sqrt{ab}.$$
Rearranging gives $$\left(\sqrt{a} - \sqrt{b}\right)^2 \geqslant 0.$$ Can you finish it now?
A: Prove that
$1)\quad\sqrt{ab}\geqslant\dfrac2{\dfrac1a+\dfrac1b}\;$  for all positive real numbers $\,a\,$ and $\,b\;.$
$2)\quad\sqrt{ab}=\dfrac2{\dfrac1a+\dfrac1b}\;\iff\;a=b\;.$
Proof :
First we will prove the inequality $\;1)\;.$
For all positive real numbers $\,a\,$ and $\,b\,,\,$ it results that
$\sqrt{ab}=\sqrt{ab}\cdot\color{blue}{\dfrac{a+b}{b+a}}\geqslant\sqrt{ab}\cdot\color{blue}{\dfrac{a+b-\left(\sqrt a-\sqrt b\right)^2}{b+a}}=$
$\!\qquad=\sqrt{ab}\cdot\dfrac{2\sqrt{ab}}{b+a}=\dfrac{2ab}{b+a}=\dfrac2{\dfrac1a+\dfrac1b}\quad,$
hence ,
$\sqrt{ab}\geqslant\dfrac2{\dfrac1a+\dfrac1b}\;$  for all positive real numbers $\,a\,$ and $\,b\;.$
Now we will prove the equivalence $\;2)\;.$
$\sqrt{ab}=\dfrac2{\dfrac1a+\dfrac1b}\;\iff\;\dfrac{a+b}{b+a}=\dfrac{a+b-\left(\sqrt a-\sqrt b\right)^2}{b+a}\;\iff$
$\qquad\qquad\qquad\,\iff\;\left(\sqrt a-\sqrt b\right)^2=0\;\iff\;a=b\quad,$
hence ,
$\sqrt{ab}=\dfrac2{\dfrac1a+\dfrac1b}\;\iff\;a=b\;.$
