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From Amir Beck book IntroductIon to nonlinear optimization p.29

We have function $f(x,y) = \frac{x+y}{x^2+y^2+1}$

"... for any $(x,y)^T \in R^2 $ , $f(x,y) = \frac{x+y}{x^2+y^2+1} <= \sqrt{2} \frac{\sqrt{x^2+y^2}}{x^2+y^2+1} <= \sqrt{2} max\frac{t}{t^2+1}$ where the first inequality follows from the Cauchy–Schwarz inequality."

Can someone explain how Cauchy-Schwarz inequality used here ?

Daulet Karim
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  • $x^2 - 2xy + y^2 = (x-y)^2 \geq 0 \implies x^2 + y^2 \geq 2xy$, $(x+y)^2 = x^2 + 2xy + y^2 \leq 2(x^2+y^2) \implies x+y \leq \sqrt{2} \sqrt{x^2+y^2}$ – User203940 Feb 25 '22 at 18:20
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    We have (by Cauchy-Schwarz) $$ x+y= 1\cdot x + 1\cdot y \leq \sqrt{1+1} \cdot \sqrt{x^2+y^2}.$$ – Severin Schraven Feb 25 '22 at 18:29
  • @SeverinSchraven thanks – Daulet Karim Feb 25 '22 at 18:31
  • You're welcome :) – Severin Schraven Feb 25 '22 at 18:31
  • @SeverinSchraven sorry but could you further explain how authors claim that "since for any t >0 , the Inequality $t^2+1 > 2t$ holds, we have that f(x,y) < $\frac{1}{\sqrt{2}}$ for any (x,y) in R^2. Thus $\frac{1}{\sqrt{2}}$ is global maximum of f(x,y)" ? I couldn't figure out how they arraived to that ? – Daulet Karim Feb 25 '22 at 18:47
  • Note that $t = 1$ gives us an equality, so it can't be strict for all $t > 0$.

    Use the fact that $(t-1)^2 = t^2 - 2t + 1 \geq 0$ for every $t \in \mathbb{R}$ to get that $t^2 + 1 \geq 2t$ (alternatively use CS). This implies that $\frac{1}{t^2+1} \leq \frac{1}{2t}$, so $\frac{t}{t^2+1} \leq \frac{t}{2t} = \frac{1}{2}$. Hence we have $f(x,y) \leq \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$ for every $(x,y) \in \mathbb{R}^2$. The maximum is achieved when $t=1$.

     – User203940 Feb 25 '22 at 18:55
  • @User203940 you are genius. Thank you!! – Daulet Karim Feb 25 '22 at 19:04

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