Am-Gm inequality

Algebra Level 3

( a + b ) ( b + c ) ( c + a ) k a b c (a+b)(b+c)(c+a)\geq kabc Find the largest positive integer k k such that the inequality above holds true for all positive numbers a , b a,b and c c .

Bonus : Determine the conditions for equality and prove it.


The answer is 8.

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1 solution

By AM-GM inequality, a + b 2 a b a+b \ge 2\sqrt{ab} , b + c 2 b c b+c \ge 2\sqrt{bc} , c + a 2 c a c+a \ge 2\sqrt{ca} .

Multiplying the 3 inequalities, we get ( a + b ) ( b + c ) ( c + a ) 8 a b c (a+b)(b+c)(c+a) \ge 8abc . Hence k = 8 k=8 .

There is equality if and only if each of three inequalities are equations.

Solving the equations we get equality for a = b = c a=b=c .

Typo: a + b 2 a b , b + c 2 b c , c + a 2 c a a+b \geq 2\sqrt{ab},\;b+c \geq 2\sqrt{bc},\;c+a \geq 2\sqrt{ca}

Hung Woei Neoh - 5 years ago

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Fixed it. Thanks.

Very simple

I Gede Arya Raditya Parameswara - 4 years, 4 months ago

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