Is No Square Negative?

Algebra Level pending

For positive reals a a and b b , find the maximum value of the expression below.

2 ( a ( a + b ) 3 + b a 2 + b 2 ) a 2 + b 2 \frac{\sqrt{2}\left(\sqrt{a(a+b)^{3}}+b\sqrt{a^{2}+b^{2}}\right)}{a^2+b^2}


The answer is 3.

Nitin Kumar by Nitin Kumar , 14, India

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1 solution

Nitin Kumar
Apr 17, 2020

2 ( a ( a + b ) 3 + b a 2 + b 2 ) \sqrt{2}(\sqrt{a(a+b)^3}+b\sqrt{a^2+b^2})

= ( 2 a 2 + 2 a b ) ( a + b ) 2 + 2 b 2 ( a 2 + b 2 ) =\sqrt{(2a^2+2ab)(a+b)^2}+\sqrt{2b^2(a^2+b^2)}

≤ ( 3 a 2 + b 2 ) ( 2 a 2 + 2 b 2 ) + 2 b 2 ( a 2 + b 2 ) \leq \sqrt{(3a^2+b^2)(2a^2+2b^2)}+\sqrt{2b^2(a^2+b^2)} because a 2 + b 2 ≥ 2 a b a^{2}+b^{2} \geq 2ab

≤ 3 a 2 + b 2 + 2 a 2 + 2 b 2 2 + 2 b 2 + a 2 + b 2 2 = 3 ( a 2 + b 2 ) . \leq \frac{3a^2+b^2+2a^2+2b^2}{2}+ \frac{2b^2+a^2+b^2}{2}= 3(a^2+b^2). because x y ≤ x + y 2 \sqrt{xy} \leq \frac{x+y}{2}

This problem was taken from the Mathematical Olympiad Summer Program- 2005 Red Group Algebra Problem 1

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