Use Cauchy

Algebra Level 4

Find the maximum value of ( x + 2 y + 3 x 2 + y 2 + 1 ) 2 \left(\dfrac{x+2y+3}{ \sqrt {x^2+y^2+1}}\right)^{2} for every x , y ∈ R x,y \in \mathbb{R} .


The answer is 14.

Bryan Lee Shi Yang by Bryan Lee Shi Yang , 17, Malaysia

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2 solutions

Steven Yuan
Mar 22, 2018

By Titu's Lemma (a form of Cauchy-Schwarz),

( x + 2 y + 3 ) 2 x 2 + y 2 + 1 ≤ x 2 x 2 + ( 2 y ) 2 y 2 + 3 2 1 = 1 + 4 + 9 = 14 . \dfrac{(x+2y+3)^2}{x^2 + y^2 + 1} \leq \dfrac{x^2}{x^2} + \dfrac{(2y)^2}{y^2} + \dfrac{3^2}{1} = 1 + 4 + 9 = \boxed{14}.

Equality occurs when x 1 = y 2 = 1 3 , \dfrac{x}{1} = \dfrac{y}{2} = \dfrac{1}{3}, or ( x , y ) = ( 1 3 , 2 3 ) . (x, y) = \left ( \dfrac{1}{3}, \dfrac{2}{3} \right ).

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We can also solve by vectors...

Arunava Das - 3 years, 2 months ago
Lucas Machado
Mar 22, 2018

Use inequality of cauchy schwartz

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