RMO 2013

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

#Integration #MathProblem #Math

Note by Gopal Chpidhary
7 years, 6 months ago

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3 votes

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Comments

Using Titu,s Lima..

a2c−1+b2d−1+c2e−1+d2a−1+e2b−1≥(a+b+c+d+e)2(a+b+c+d+e)−5\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}

jagdish singh - 7 years, 6 months ago

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.

Finn Hulse - 7 years, 6 months ago

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

gopal chpidhary - 7 years, 6 months ago

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.

easha manideep d - 5 years, 6 months ago

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.

Siddharth Kumar - 7 years, 6 months ago
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