Inequalities are Amazing!

Algebra Level pending

Find the largest y such that 1 1 + x 2 y x y + x \frac{1}{1+x^2}\geq \frac{y-x}{y+x} for all x > 0 x>0


The answer is 2.828.

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1 solution

James Hennessy
Oct 28, 2016

1 1 + x 2 y x y + x \frac {1}{1 + x^2} \geq \frac {y - x}{y + x}

( y + x ) ( y x ) ( 1 + x 2 ) (y + x) \geq (y - x)(1 + x^2)

y + x y x + x 2 y x 3 y + x \geq y - x +x^2y - x^3

2 x x 2 y x 3 2x \geq x^2y - x^3

x y 2 + x 2 xy \leq 2 + x^2

y 2 x + x y \leq \frac {2}{x} + x

Define a function f ( x ) = 2 x + x f(x)= \frac {2}{x} + x

f ( x ) = 2 x 2 + 1 f'(x)=-2x^{-2} + 1

2 x 2 + 1 = 0 -2x^{-2} + 1 = 0

x = ± 2 x= \pm \sqrt{2}

f ( x ) = 4 x 3 f''(x)=4x^{-3}

f ( 2 ) = 4 ( 2 ) 3 > 0 f''(\sqrt{2})= \frac {4}{(\sqrt{2})^3} >0

Therefore the minimum value of x when x>0 is 2 \sqrt{2} .

So y 2 2 + 2 y \leq \frac {2}{\sqrt{2}} + \sqrt{2}

y = 2 2 = 0.282... y = 2 \sqrt{2} = 0.282...

Just wanted to practice my latex haha

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