Inspired by Alan Yan

Algebra Level 2

If a , b , c a,b,c are positive reals such that a b c = 1 abc=1 , find the minimum value of the following expression:

a 2 ( b + c ) + b 2 ( c + a ) + c 2 ( a + b ) \large a^2(b+c)+b^2(c+a)+c^2(a+b)

Inspiration .


The answer is 6.00.

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Nihar Mahajan by Nihar Mahajan , 20, India

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2 solutions

Alan Yan
Sep 27, 2015

∑ s y m a 2 b ≥ 6 a b c 6 = 6 \sum_{sym}{a^2b} \geq 6\sqrt[6]{abc} = 6 By AM-GM.

Equality a = b = c = 1 a = b = c = 1 .

7 Helpful 0 Interesting 0 Brilliant 0 Confused

Ha , that was nice.

Nihar Mahajan - 5 years, 8 months ago

Exactly the same way

Raushan Sharma - 5 years, 8 months ago

I think it's apt for level one

Bhaskar Pandey - 3 years, 8 months ago
Nihar Mahajan
Sep 27, 2015

∑ c y c a 2 ( b + c ) = a b c ( ∑ c y c a ( b + c ) b c ) = ∑ c y c ( a b + b a ) ≥ 6 ∵ a b + b a ≥ 2 ( b y A M − G M ) \sum_{cyc} a^2(b+c) = abc\left(\sum_{cyc} \dfrac{a(b+c)}{bc}\right) =\sum_{cyc} \left(\dfrac{a}{b}+\dfrac{b}{a}\right) \geq \boxed{6} \because \dfrac{a}{b} + \dfrac{b}{a} \geq 2 \ ( \ by \ AM-GM )

Equality iff a = b = c = 1 a=b=c=1 .

4 Helpful 0 Interesting 0 Brilliant 0 Confused

I did using calculus. Taking minima.

Aditya Kumar - 5 years, 8 months ago

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Post it as a separate solution :)

Nihar Mahajan - 5 years, 8 months ago

How bro explain in brief

Naman Anand - 5 years, 7 months ago

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