Standard Inequality

Algebra Level 3

Let a a , b b , c c be positive reals such that 1 a + 1 b + 1 c = 1 \frac{1}{a} +\frac{1}{b}+\frac{1}{c} =1 . Find the minimum value of ( a 1 ) ( b 1 ) ( c 1 ) (a-1)(b-1)(c-1) .


The answer is 8.

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1 solution

Mohammed Imran
Apr 3, 2020

Let f ( x ) = 1 x x f(x)=\frac{1-x}{x} . Since f ( x ) f(x) is convex, applying Jensen's Inequality, we have f ( 1 a + 1 b + 1 c 3 ) f ( 1 a ) + f ( 1 b ) + f ( 1 c ) 3 f(\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}) \leq \frac{f(\frac{1}{a})+f(\frac{1}{b})+f(\frac{1}{c})}{3} this reduces to ( a 1 ) + ( b 1 ) + ( c 1 ) 6 (a-1)+(b-1)+(c-1) \geq 6 . Since a , b , c > 1 a,b,c>1 , applying A.M-G.M inequality, we have ( a 1 ) ( b 1 ) ( c 1 ) 8 (a-1)(b-1)(c-1) \geq \boxed{8}

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