Question

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

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

Solution

## The correct option is A Greater than or equal to $$10$$$$a,b,c$$ in H.P, $$b=\dfrac{2ac}{a+c}...(1)$$$$\dfrac{3a+2b}{2a-b}+\dfrac{3c+2b}{2c-b}=\dfrac{6ac+4bc-3ab-2b^2+6ac+4ab-3bc-2b^2}{4ac-2ab-2bc+b^2}$$$$=\dfrac{12ac+b(a+c)-4b^2}{4ac-2b(a+c)+b^2}$$Replace $$b(a+c)=2ac$$ by relation (1)$$=\dfrac{12ac+2ac-4b^2}{4ac-4ac+b^2}=\dfrac{14ac-4b^2}{b^2}$$$$=\dfrac{14ac}{b^2}-4$$Replace $$\dfrac{2ac}{b}=a+c$$,$$\dfrac{7(a+c)}{b}-4$$As $$a,b,c$$ in H.P $$\therefore \dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}$$$$\Rightarrow \dfrac{7(a+c)}{b}-4=7(a+c)\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{c}\right)-4$$$$=\dfrac{7}{2}\left(1+\dfrac{a}{c}+\dfrac{c}{a}+1\right)-4$$$$=\dfrac{7}{2}\left(2+\dfrac{a}{c}+\dfrac{c}{a}\right)-4=3+\dfrac{7}{2}\left(\dfrac{a}{c}+\dfrac{c}{a}\right)$$By Arithmetic mean relation on $$\dfrac{a}{c},\dfrac{c}{a}$$$$\Rightarrow \dfrac{\left(\dfrac{a}{c}+\dfrac{c}{a}\right)}{2}\geq \sqrt{\dfrac{a}{c}\times\dfrac{c}{a}}$$$$\dfrac{a}{c}+\dfrac{c}{a}\geq 2$$$$\therefore 3+\dfrac{7}{2}\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\geq 3+\dfrac{7}{2}\times 2=10$$Will be greater than or equal to 10.Mathematics

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