A Useful Inequality

$\begin{aligned} \frac 1{\frac 1a + \frac 1b} & \le \frac {a+b}2 \\ \frac {ab}{a+b} & \le \frac {a+b}2 \\ 2ab & \le (a+b)^2 \\ (a+b)^2 & \ge 2ab \\ a^2 + b^2 + 2ab & \ge 2ab \\ a^2+b^2 & \ge 0 & \small \color{#3D99F6}{\text{Since } a > 0 \text{ and }b>0} \\ \implies a^2+b^2 & > 0 \end{aligned}$ .

Therefore, the inequality is true for $\boxed{\text{all of them}}$ ( $a$ and $b$ ).

By rearranging, the inequality can be seen to be equivalent to $(a+b)^2 \geq 2ab$ that is clearly true since: $(a+b)^2 = a^2+b^2+2ab \geq 2ab$ .

The inequality is a special case of the AM - HM inequality, which statest that the Arithmetic Mean is always greater or equal than the Harmonic Mean.