Minimum of Squares!

Algebra Level 3

If a + b = 1 a n d a , b > 0 \color{#0C6AC7}a + \color{#456461}b = \color{magenta}{1} \quad and \quad \color{#0C6AC7}a , \color{#456461}b > \color{#624F41}0
Then minimum value \text{minimum value} of :

( a + 1 a ) 2 + ( b + 1 b ) 2 ≥ ? \left( \color{#0C6AC7}a \color{#EC7300}+ \dfrac{\color{magenta}1}{\color{#0C6AC7}a} \right)^{\color{#D61F06}2 }\color{#3D99F6}+ \left( \color{#456461}b \color{#EC7300}+ \dfrac{\color{magenta}1}{\color{#456461}b} \right)^{\color{#D61F06}2} \geq \ ?


The answer is 12.5.

Akhil Bansal by Akhil Bansal , 22, India

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1 solution

Hoàn Phạm
Oct 14, 2015

P = ( a + 1 a ) 2 + ( b + 1 b ) 2 = a 2 + b 2 + 1 a 2 + 1 b 2 + 4 = ( a 2 + b 2 ) ( 1 + 1 a 2 b 2 ) + 4 P = \left( a+\dfrac{1}{a}\right)^2 + \left( b+\dfrac{1}{b}\right)^2 = a^2+b^2 + \dfrac{1}{a^2}+\dfrac{1}{b^2} + 4 = \left(a^2+b^2\right)\left(1+\dfrac{1}{a^2 b^2}\right) + 4 We have a 2 + b 2 ≥ ( a + b ) 2 2 = 1 2 a^2+b^2 \geq \dfrac{(a+b)^2}{2} = \dfrac{1}{2} and, a b ≤ ( a + b 2 ) 2 = 1 4 ab \leq \left(\dfrac{a+b}{2}\right)^2= \dfrac{1}{4} So, P ≥ 1 2 ( 1 + 16 ) + 4 = 25 2 P \geq \dfrac{1}{2} \left( 1+ 16\right)+4 = \dfrac{25}{2} IFF a = b = 1 2 a = b = \dfrac{1}{2} the equation happens

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