If a, b and c are three positive numbers in an arithmetic progression, then:

Since, a, b, c, are in AP

$\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in H.P

$\Rightarrow \frac{a+c}{2}\ge \frac{2ac}{a+c}$ $\left[Since , AM\ge \mathrm{HM}=\frac{2\mathrm{ac}}{\mathrm{a}+\mathrm{c}}\right]$

$\Rightarrow b\ge \frac{2ac}{a+c}$   

$\Rightarrow ab+bc\ge 2ac$