Question

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Answer 1

Given Question :-

Prove that,

$\sf \:\dfrac{2}{a + b} + \dfrac{2}{b + c} + \dfrac{2}{c + a} < \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \: where \: a,b,c > 0$

$\green{\large\underline{\sf{Solution-}}}$

Given that,

$\rm :\longmapsto\:a,b,c > 0$

$\rm \implies\:\dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c} > 0$

We know,

Arithmetic mean and Harmonic mean between two distinct real numbers are interrelated by the relationship

$\boxed{\tt{ Arithmetic \: mean > Harmonic \: mean}}$

Now,

$\rm :\longmapsto\:\sf \: Let \: us \: consider \: two \: numbers \: \dfrac{1}{a} \: and \: \dfrac{1}{b}$

So,

$\rm :\longmapsto\:\dfrac{1}{2} \bigg[\dfrac{1}{a} + \dfrac{1}{b} \bigg] > \dfrac{2}{\dfrac{1}{\dfrac{1}{a} } + \dfrac{1}{\dfrac{1}{b} } }$

$\rm \implies\:\dfrac{1}{a} + \dfrac{1}{b} > \dfrac{4}{a + b} - - - - (1)$

Similarly,

$\rm \implies\:\dfrac{1}{b} + \dfrac{1}{c} > \dfrac{4}{b + c} - - - - (2)$

and

$\rm \implies\:\dfrac{1}{c} + \dfrac{1}{a} > \dfrac{4}{c + a} - - - - (3)$

On adding equation (1), (2) and (3), we get

$\rm :\longmapsto\:2\bigg[\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \bigg] > \dfrac{4}{a + b} + \dfrac{4}{b + c} + \dfrac{4}{c + a}$

$\rm :\longmapsto\:\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} > \dfrac{2}{a + b} + \dfrac{2}{b + c} + \dfrac{2}{c + a}$

$\bf\implies \:\dfrac{2}{a + b} + \dfrac{2}{b + c} + \dfrac{2}{c + a} < \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}$

HENCE , PROVED

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Additional Information

Let us consider two positive real numbers a and b, then

Arithmetic mean between a and b is

$\rm \implies\:\boxed{\tt{ \: AM = \frac{a + b}{2} \: }}$

Geometric mean between a and b is

$\rm \implies\:\boxed{\tt{ \: GM = \sqrt{ab} \: }}$

Harmonic mean between a and b is

$\rm \implies\:\boxed{\tt{ \: HM = \frac{2}{ \dfrac{1}{a} + \dfrac{1}{b} } \: }}$

$\red{\rm :\longmapsto\:\boxed{\tt{ AM \geqslant GM \geqslant HM \: }}}$