Hey guys check this out.
It’s a rmo 2013 question
Given any real numbers a,b,c,d,e>1, Prove that
a^2/(c-1) + b^2/(d-1) +c^2/(e-1) +d^2/(a-1) +e^2/(b-1) ≥ 20

Apply AM GM inequality I mean put 20 as 4.5 and send 5 that side. Hence it becomes AM and from then You'll get a^2/a-1 is greater than or equal to 4 you can prove it by saying (a-2)^2 is greater than or equal to 0.

Because you tagged this "integration", then I must assume it is the simplified form of an integral, and that you can geometrically represent this area and have it never be less than 20 within its given bounds. But that's just a guess.

I first substituted a=1+v, b=1+w, c=1+x, d=1+y, e=1+z where v,w,x,y,z are non negative real numbers. Then apply rearrangement inequality, and the final answer would be derived from AM-G.M. Just try it yourself.

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TopNewestUsing Titu,s Lima..

\(\displaystyle \frac{a^2}{c-1}+\frac{b^2}{d-1}+\frac{c^2}{e-1}+\frac{d^2}{a-1}+\frac{e^2}{b-1}\geq \frac{(a+b+c+d+e)^2}{(a+b+c+d+e)-5}\)

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Apply AM GM inequality I mean put 20 as 4.5 and send 5 that side. Hence it becomes AM and from then You'll get a^2/a-1 is greater than or equal to 4 you can prove it by saying (a-2)^2 is greater than or equal to 0.

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Because you tagged this "integration", then I must assume it is the simplified form of an integral, and that you can geometrically represent this area and have it never be less than 20 within its given bounds. But that's just a guess.

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no, you can't think that way... it will become complicated.

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I first substituted a=1+v, b=1+w, c=1+x, d=1+y, e=1+z where v,w,x,y,z are non negative real numbers. Then apply rearrangement inequality, and the final answer would be derived from AM-G.M. Just try it yourself.

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