Question

The sum of two nonzero real numbers is $6.$ Then the minimum value of the sum of their reciprocals is ?

Answer 1

Let the two nonzero real numbers be $a,b$

Given $a+b=6$

$AM\ge GM\ge HM⟹\frac{a+b}{2}\ge \sqrt{ab}\ge \frac{2}{\frac{1}{a}+\frac{1}{b}}$

$⟹\frac{1}{a}+\frac{1}{b}\ge \frac{2}{3}$

Minimum value of the sum of their reciprocals is $\frac{2}{3}$