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I came across the following inequality as a step in the proof of another: Let $x,\ y,\ z>0$, then this is true: $$\sqrt{\frac{x^2 + 1}{y^2 + 1}} + \sqrt{\frac{y^2 + 1}{z^2 + 1}} + \sqrt{\frac{z^2 + 1}{x^2 + 1}}\le \frac{x}{y} + \frac{y}{z} + \frac{z}{x}.$$ I checked it numerically and it seems to be valid, but I don't see an obvious solution. For only two variables, it is easy to show the following: $$\sqrt{\frac{x^2+\delta}{y^2+\delta}} + \sqrt{\frac{y^2+\delta}{x^2 + \delta}} \le \frac{x^2 + y^2 + 2\delta}{\sqrt{x^2y^2+2xy\delta + \delta^2}} = \frac{x^2 + y^2 + 2\delta}{xy + \delta}\le \frac{x^2+y^2}{xy} $$ since the left hand side of the last inequality is a decreasing function. However, I cannot extend this logic to the full inequality. Any help is appreciated.

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  • $\begingroup$ What you mean by $\sum_{cyc}$? $\endgroup$ – Qurultay Jun 18 '18 at 20:37
  • $\begingroup$ I clarified it a bit, thanks. $\endgroup$ – Ivan Jun 18 '18 at 20:39
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Let $z=\min\{x,y,z\}$.

Thus, we need to prove that $$\frac{x}{y}+\frac{y}{x}-2+\frac{y}{z}-\frac{y}{x}+\frac{z}{x}-1\geq\sqrt{\tfrac{x^2+1}{y^2+1}}+\sqrt{\tfrac{x^2+1}{y^2+1}}-2+\sqrt{\tfrac{y^2+1}{z^2+1}}-\sqrt{\tfrac{y^2+1}{x^2+1}}+\sqrt{\tfrac{z^2+1}{x^2+1}}-1$$ or $$\tfrac{(x-y)^2}{xy}+\tfrac{(z-x)(z-y)}{xz}\geq\tfrac{\left(\sqrt{x^2+1}-\sqrt{y^2+1}\right)^2}{\sqrt{(x^2+1)(y^2+1)}}+\tfrac{\left(\sqrt{z^2+1}-\sqrt{x^2+1}\right)\left(\sqrt{z^2+1}-\sqrt{y^2+1}\right)}{\sqrt{(x^2+1)(z^2+1)}},$$ which is true because $$(x-y)^2\geq\left(\sqrt{x^2+1}-\sqrt{y^2+1}\right)^2,$$ $$x-z\geq\sqrt{x^2+1}-\sqrt{z^2+1},$$ $$y-z\geq\sqrt{y^2+1}-\sqrt{z^2+1},$$ $$\sqrt{(x^2+1)(y^2+1)}\geq xy$$ and $$\sqrt{(x^2+1)(z^2+1)}\geq xz.$$ For example, $$x-z\geq\sqrt{x^2+1}-\sqrt{z^2+1}$$ it's $$x-z\geq\frac{x^2-z^2}{\sqrt{x^2+1}+\sqrt{z^2+1}}$$ or $$\sqrt{x^2+1}+\sqrt{z^2+1}\geq x+z,$$ which is obvious.

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  • $\begingroup$ Thanks, very nice! $\endgroup$ – Ivan Jun 18 '18 at 20:57
  • $\begingroup$ You are welcome! $\endgroup$ – Michael Rozenberg Jun 18 '18 at 20:57

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