Problem 16

Algebra Level 5

4 a + 3 b + c 3 ( a − b ) b \large 4a+3b+\dfrac{c^3}{(a-b)b}

Given that a , b a,b and c c are positive reals satisfying a > b a> b and a + b + c = 4 a+b+c=4 , find the minimum value of the expression above.

This problem is in this Set .


The answer is 12.

Son Nguyen by Son Nguyen , 20, Vietnam

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1 solution

P C
Feb 27, 2016

The expression can be written as 3 ( a + b ) + a − b + b + c 3 b ( a − b ) 3(a+b)+a-b+b+\frac{c^3}{b(a-b)} Applying AM-GM and we have 3 ( a + b ) + a − b + b + c 3 b ( a − b ) ≥ 3 ( a + b + c ) = 12 3(a+b)+a-b+b+\frac{c^3}{b(a-b)}\geq 3(a+b+c)=12

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