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Let $x,y,z>0$. Then $$\sum_\text{cyc} \frac{x}{\sqrt{x^2+y^2}}>1$$

I found a similar inequality in the other direction but I can‘t apply Cauchy-Schwarz here... All I see is by Cauchy-Schwarz,

$$\sum_\text{cyc} \frac{x}{\sqrt{x^2+y^2}}\geq \frac{\sum_\text{cyc}\sqrt x}{\sum_\text{cyc}\sqrt[4]{x^2+y^2}}$$ which is not helpful.

Bernard
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1 Answers1

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Note that $$\sum_{\rm cyc} \frac x{\sqrt{x^2+y^2}}>\sum_{\rm cyc} \frac x{x+y}$$ $$>\sum_{\rm cyc}\frac x{x+y+z}=1.$$

Pythagoras
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