Question

If $a,b,c$ are distinct positive numbers in H.P. and if $\frac{a+b}{2a-b}+\frac{c+b}{2c-b}\ge \sqrt{\lambda \sqrt{\lambda \sqrt{\lambda \cdots \infty }}}$, then find the minimum integral value of $\lambda$.

Answer 1

If $a,b,c$ are distinct positive:
$p=\frac{a+b}{2a-b}+\frac{c+b}{2c-b}$

$=\frac{a+\frac{2ac}{a+c}}{2a-\frac{2ac}{a+c}}+\frac{c+\frac{2ac}{a+c}}{2c-\frac{2ac}{a+c}}$ ....... $[\because a,b,c$ are in H.P.$]$

$=\frac{a^2+3ac}{2a^2}+\frac{3ac+c^2}{2c^2}$

$=\frac{a+3c}{2a}+\frac{3a+c}{2c}=1+\frac{3}{2}(\frac{c}{a}+\frac{a}{c})\ge 4$

Also $\sqrt{\lambda \sqrt{\lambda \sqrt{\lambda .....\infty }}}={\lambda }^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots }$

Let $S=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$

Using summation of GP, we get

$S=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1$

Also $\sqrt{\lambda \sqrt{\lambda \sqrt{\lambda .\infty }}}={\lambda }^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots }=\lambda$

$\therefore \lambda \ge 4$