segunda-feira, 23 de janeiro de 2012

DESIGUALDADES


how to prove that (a+b)(b+c)(c+a)>8abc



First we prove the lemma: 

a%2Bb+%3E=+2sqrt%28ab%29

Proof:
%28a-b%29%5E2%3E=0       <----because the square of a real number is 
                            never negative

a%5E2-2ab%2Bb%5E2%3E=0   <----squaring out the left side

a%5E2%2B2ab%2Bb%5E2%3E=4ab <----adding 4ab to both sides

%28a%2Bb%29%5E2%3E=4ab     <----factoring the left side 

%28a%2Bb%29%3E=2sqrt%28ab%29     <----taking non-negative square roots
                            of both sides

Similarly we can prove 

%28b%2Bc%29%3E=2sqrt%28bc%29

and

%28c%2Ba%29%3E=2sqrt%28ca%29

just by changing the letters in the first proof.

Therefore, multiplying the left and right sides of the
three inequalities:

system%28%28a%2Bb%29%3E=2sqrt%28ab%29%2C%28b%2Bc%29%3E=2sqrt%28bc%29%2C%28c%2Ba%29%3E=2sqrt%28ac%29%29

%28a%2Bb%29%28b%2Bc%29%28c%2Ba%29%3E=%282sqrt%28ab%29%29%282sqrt%28bc%29%29%282sqrt%28ca%29%29=8sqrt%28a%5E2b%5E2c%5E2%29=8abc



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