Question

If a, b, c are in H.P., then $\frac{a+b}{2a-b}+\frac{c+b}{2c-b}>4$

• A. True
• B. False

Answer 1

The correct option is A True

As a, b, c are in H.P., $\therefore b=\frac{2ac}{a+c}$
$\frac{a+b}{2a-b}=\frac{a+\frac{2ac}{a+c}}{2a-\frac{2ac}{a+c}}$
$=\frac{a^2+3ac}{2a^2}=\frac{1}{2}+\frac{3}{2}.\frac{c}{a}$
Similarly $\frac{c+b}{2c-b}=\frac{1}{2}+\frac{3}{2}.\frac{a}{c}$
Hence we have to prove that
$\frac{1}{2}+\frac{3}{2}[\frac{c}{a}+\frac{a}{c}]+\frac{1}{2}>4$
or $\frac{3}{2}[\frac{c}{a}+\frac{a}{c}]>3$ or $\frac{c^2+a^2}{2ac}>1$
or $\frac{c^2+a^2}{2}>ac$ which is true $\because A.M>G.M$