Question

$a,b,c$ are positive real numbers which are in HP. then $\frac{(a+b)}{(2a-b)}+\frac{(c+b)}{(2c-b)}$ is

• A. $\ge 4$
• B. $0$
• C. $\le 4$
• D. None of these

Answer 1

The correct option is A

$\ge 4$

Explanation of the correct option:

Solve the given expression using the condition of H.P:

$\frac{(a+b)}{(2a-b)}+\frac{(c+b)}{(2c-b)}$ …$\left(i\right)$

$\because$$a,b,c$ are in HP.

Then $\mathbf{b}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{2}\mathbf{ac}}{\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{c}\mathbf{)}}$

Substitute the value of $b$ in equation $\left(i\right)$:

$\frac{(a+b)}{(2a-b)}+\frac{(c+b)}{(2c-b)}$$=\frac{a+\left({\displaystyle \frac{2ac}{a+c}}\right)}{2a-\left({\displaystyle \frac{2ac}{a+c}}\right)}+\frac{c+\left({\displaystyle \frac{2ac}{a+c}}\right)}{2c-\left({\displaystyle \frac{2ac}{a+c}}\right)}$

By solving above equation we get $1+\frac{3}{2}\left(\frac{c}{a}+\frac{a}{c}\right)$

$=1+\frac{3}{2}\left[\frac{c}{a}+\frac{a}{c}\right]$

$\Rightarrow$$1+\frac{3}{2}\left[\frac{c}{a}+\frac{a}{c}\right]$$\ge 1+\frac{3}{2}\left[2\right]$ $\left[\because \mathrm{AM}\ge \mathrm{GM}\right]$

$\Rightarrow$$1+\frac{3}{2}\left[\frac{c}{a}+\frac{a}{c}\right]$$\ge 4$

Hence, option ‘A’ is correct.