Question-175396

Question Number 175396 by Shrinava last updated on 29/Aug/22

Answered by mahdipoor last updated on 29/Aug/22

x + y^{2} \sqrt{\frac{2}{x^{2} + y^{2}}} ⩾ y + \sqrt{\frac{x^{2} + y^{2}}{2}} \Leftrightarrow

(x + y^{2} \sqrt{\frac{2}{x^{2} + y^{2}}}) (\sqrt{\frac{x^{2} + y^{2}}{2}}) ⩾ (y + \sqrt{\frac{x^{2} + y^{2}}{2}}) (\sqrt{\frac{x^{2} + y^{2}}{2}}) \Leftrightarrow

x \sqrt{\frac{x^{2} + y^{2}}{2}} + y^{2} ⩾ y \sqrt{\frac{x^{2} + y^{2}}{2}} + \frac{x^{2} + y^{2}}{2} \Leftrightarrow

(x - y) \sqrt{\frac{x^{2} + y^{2}}{2}} ⩾ \frac{x^{2} - y^{2}}{2} = \frac{(x - y) (x + y)}{2} \overset{I}{\Leftrightarrow}

\sqrt{\frac{x^{2} + y^{2}}{2} ⩾} \frac{x + y}{2} \overset{II}{\Leftrightarrow} \frac{x^{2} + y^{2}}{2} ⩾ \frac{x^{2} + y^{2} + 2 x y}{4} \Leftrightarrow

\frac{x^{2} + y^{2} - 2 x y}{4} = \frac{{(x - y)}^{2}}{4} ⩾ 0

I : x - y ⩾ 0 o r x ⩾ y

II : x + y ⩾ 0

Answered by MJS_new last updated on 29/Aug/22

let y = p x with p > 0

x \frac{\sqrt{2} p^{2} + \sqrt{p^{2} + 1}}{\sqrt{p^{2} + 1}} ⩾ x \frac{\sqrt{2} p + \sqrt{p^{2} + 1}}{\sqrt{2}}

\frac{\sqrt{2} p^{2} + \sqrt{p^{2} + 1}}{\sqrt{p^{2} + 1}} ⩾ \frac{\sqrt{2} p + \sqrt{p^{2} + 1}}{\sqrt{2}}

2 p^{2} + \sqrt{2 (p^{2} + 1)} ⩾ p^{2} + 1 + p \sqrt{2 (p^{2} + 1)}

(p - 1) (p + 1 - \sqrt{2 (p^{2} + 1)}) ⩾ 0

only true for 0 < p ⩽ 1

\Rightarrow only true for 0 < y ⩽ x

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