Question

If positive numbers $a,b,c$ are in HP and if $\left(\frac{a+b}{2a-b}\right)+\left(\frac{c+b}{2c-b}\right)\ge \sqrt{\lambda \sqrt{\lambda \sqrt{\lambda .....\infty }}}$ then find maximum value of $\lambda$ such that above relation always hold?

Answer 1

$P=\frac{a+b}{2a-b}+\frac{c+b}{2c-b}$

as $a,b,c$ are in HP,

Therefore, $b=\frac{2ac}{a+c}$

Now,

$P=\frac{a+\frac{2ac}{a+c}}{2a-\frac{2ac}{a+c}}+\frac{c+\frac{2ac}{a+c}}{2c-\frac{2ac}{a+c}}$

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

Therefore,

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