Inequality proof $\frac{2 \cdot a \cdot b }{a+b} < \frac {a+b}{2}$ This is very simple, but impossible to Google as you can understand.
I need to prove this inequality:
$$\frac{2 \cdot a \cdot b }{a+b} < \frac {a+b}{2}$$
where $ a < b$ and $a > 0$ and $b > 0$
I have tried to rework it to:
$$4 \cdot a \cdot b < {(a+b)}^2$$ but that doesn't bring me further.
Also I tried to substitute $ a = b-\delta$ and substitute that in the inequality.
I can see where it goes when $a=b$ and when $b >>a$ but that still is not the proof.

We know that $(a-b)^2 > 0$ or $a^2 + b^2 -2ab  > 0$ 
Adding $2ab$ term on both sides of the inequality we get 
$a^2 + b^2 +2ab - 2ab > 2ab$ 
$a^2 + b^2 + 2ab > 4ab$
$(a+b)^2 > 4ab$
Equivalently
$\frac{(a+b)}{2} > \frac{2(ab)}{a+b}$
