Question

If $a,b,c$ are three distinct positive real numbers which are in H.P then $\frac{3a+2b}{2a-b}+\frac{3c+2b}{2c-b}$ is

• A. Greater than or equal to $10$
• B. Less than or equal to $10$
• C. Only equal to $20$
• D. None of these

Answer 1

The correct option is A Greater than or equal to $10$

$a,b,c$ in H.P, $b=\frac{2ac}{a+c}...\left(1\right)$

$\frac{3a+2b}{2a-b}+\frac{3c+2b}{2c-b}=\frac{6ac+4bc-3ab-2b^2+6ac+4ab-3bc-2b^2}{4ac-2ab-2bc+b^2}$

$=\frac{12ac+b(a+c)-4b^2}{4ac-2b(a+c)+b^2}$

Replace $b(a+c)=2ac$ by relation (1)

$=\frac{12ac+2ac-4b^2}{4ac-4ac+b^2}=\frac{14ac-4b^2}{b^2}$

$=\frac{14ac}{b^2}-4$

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

$\frac{7(a+c)}{b}-4$

As $a,b,c$ in H.P $\therefore \frac{2}{b}=\frac{1}{a}+\frac{1}{c}$

$\Rightarrow \frac{7(a+c)}{b}-4=7(a+c)\frac{1}{2}(\frac{1}{a}+\frac{1}{c})-4$

$=\frac{7}{2}(1+\frac{a}{c}+\frac{c}{a}+1)-4$

$=\frac{7}{2}(2+\frac{a}{c}+\frac{c}{a})-4=3+\frac{7}{2}(\frac{a}{c}+\frac{c}{a})$

By Arithmetic mean relation on $\frac{a}{c},\frac{c}{a}$

$\Rightarrow \frac{(\frac{a}{c}+\frac{c}{a})}{2}\ge \sqrt{\frac{a}{c}\times \frac{c}{a}}$

$\frac{a}{c}+\frac{c}{a}\ge 2$

$\therefore 3+\frac{7}{2}(\frac{a}{c}+\frac{c}{a})\ge 3+\frac{7}{2}\times 2=10$

Will be greater than or equal to 10.