AM-GM Inequality?

Algebra Level 4

If a a and b b are two positive real numbers such that a + b = 1 a+b=1 , then what is the minimum value of ( a + 1 a ) 2 + ( b + 1 b ) 2 \left (a + \frac{1}{a}\right)^{2}+\left(b + \frac{1}{b}\right)^{2}


The answer is 12.5.

Yash Jain by Yash Jain , 21, India

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1 solution

Mohammed Imran
Apr 3, 2020

Let f ( x ) = x + 1 x f(x)=x+\frac{1}{x} . Since f ( x ) f(x) is a convex function, by Jensen's Inequality, we have f ( a + b 2 ) ≤ f ( a ) + f ( b ) 2 f(\frac{a+b}{2}) \leq \frac{f(a)+f(b)}{2} so, the minimum value of the above expression is 12.5 \boxed{12.5}

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