Minimum value of equations #2

Algebra Level 3

Positive real numbers a a and b b are such that a + b = 1 a+b=1 . Find the minimum value of the following expression.

( 1 + 1 a ) 2 + ( 1 + 1 b ) 2 \left(1+\frac 1a \right)^{2}+\left(1+\frac 1b\right)^{2}

Bonus: Find the solution a , b a, b so that the value of A A is minimum.


The answer is 18.

Tin Le by Tin Le , 15, Vietnam

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2 solutions

Chew-Seong Cheong
May 18, 2019

By Titu's lemma , we have:

( 1 + 1 a ) 2 + ( 1 + 1 b ) 2 ≥ 1 2 ( 2 + 1 a + 1 b ) 2 By AM-HM inequality: 2 1 a + 1 b ≤ a + b 2 = 1 2 ⟹ 1 a + 1 b ≥ 4 ≥ ( 2 + 4 ) 2 2 = 18 \begin{aligned} \left(1+\frac 1a\right)^2 + \left(1+\frac 1b\right)^2 & \ge \frac12 \left(2+\color{#3D99F6} \frac 1a + \frac 1b \right)^2 & \small \color{#3D99F6} \text{By AM-HM inequality: }\frac 2{\frac 1a+\frac 1b} \le \frac {a+b}2 = \frac 12 \implies \frac 1a + \frac 1b \ge 4 \\ & \ge \frac {(2+4)^2}2 = \boxed {18} \end{aligned}

Equality occurs when a = b = 1 2 a=b = \frac 12 .

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Dan Czinege
May 17, 2019

I think that you should add that a,b are positive real numbers, because otherwise it could have any value. (Atleast I think so. :D)

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