First Problem in 2018

Algebra Level 3

For positive reals x x and y y , is it true that

1 ( 1 + x ) 2 + 1 ( 1 + y ) 2 ≥ 2 x + y + 2 \large \dfrac{1}{(1 + \sqrt{x}) ^2} + \dfrac{1}{(1 + \sqrt{y})^2} \geq \dfrac{2}{x+y+2}

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Fidel Simanjuntak by Fidel Simanjuntak , 19, Indonesia

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1 solution

Chew-Seong Cheong
Jan 11, 2018

Relevant wiki: Titu's Lemma

By Cauchy-Schwarz inequality ( 1 + x ) 2 ≤ 2 ( 1 + x ) (1+\sqrt x)^2 \le 2(1+x) , ⟹ 1 ( 1 + x ) 2 ≥ 1 2 ( 1 + x ) \implies \dfrac 1{(1+\sqrt x)^2} \ge \dfrac 1{2(1+x)} . Therefore, we have:

1 ( 1 + x ) 2 + 1 ( 1 + y ) 2 ≥ 1 2 ( 1 + x ) + 1 2 ( 1 + y ) By Titu’s lemma ≥ ( 1 + 1 ) 2 2 + 2 x + 2 + 2 y = 2 x + y + 2 \begin{aligned} \frac 1{(1+\sqrt x)^2} + \frac 1{(1+\sqrt y)^2} & \ge \color{#3D99F6} \frac 1{2(1+x)} + \frac 1{2(1+y)} & \small \color{#3D99F6} \text{By Titu's lemma} \\ & \ge \color{#3D99F6} \frac {(1+1)^2}{2+2x+2+2y} \\ & = \boxed{\dfrac 2{x+y+2}} \end{aligned}

Equality occurs when x = y = 1 x=y=1 . Therefore, the inequality is true .

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