IIT JEE , AIEEE Forum - Mathematics , Algebra

dream_iit

if a,b,c are +ive fractions and a+ b+ c = 1 , prove

[ 1 + (1/a)] [1 + (1/b)] [1 + (1/c)]

≥ 64

hsbhatt

Prove what?

dream_iit

[ 1 + (1/a)] [1 + (1/b)] [1 + (1/c)] >= 64 ,sir its written there , by mistake it went to next line

hsbhatt

Put a = 1/3 = b = c and check. Its wrong

hsbhatt

I am guilty of being hasty and so made a calculation mistake which I realised later but could not rectify it earlier.

We are asked to prove that $\left(1+ \frac{1}{a} \right) \left(1+ \frac{1}{b} \right) \left(1+ \frac{1}{c} \right) \ge 64$ where a+b+c = 1 and all are positive.

Now, consider the function $f(x) = \log \frac{1+x}{x} = \log (1+x) - \log x$

So, that f(x) is a convex function and so we can apply Jensen's Inequality as follows:

$\frac{\log \left(1+\frac{1}{a}\right) + \log \left(1+\frac{1}{b}\right)+ \log \left(1+\frac{1}{c}\right)}{3} \ge \log \left(1 + \frac{1}{\frac{a+b+c}{3}} \right) = \log 4$ (since a+b+c = 1)

So we get $\left(1+\frac{1}{a}\right) \left(1+\frac{1}{b}\right) \left(1+\frac{1}{c}\right)} \ge 4^3 = 64$

I hope I have redeemed myself :)

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