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On the wikipedia page on "Nesbit's inequality", the fifth proof ends as follows:

$$ \frac{x+z}{y}+\frac{y+z}{x}+\frac{x+y}{z}\geq 6$$ which is true, by AM-GM inequality.

I am wondering if the inequality is obvious / immediate from just looking at it and how you see this immediately without resorting to something like the following proof: \begin{align*} \frac{x+z}{y}+\frac{y+z}{x}+\frac{x+y}{z} &\geq 2\cdot\left(\frac{\sqrt{xz}}{y}+\frac{\sqrt{yz}}{x}+\frac{\sqrt{xy}}{z}\right) \\ &\geq 6\cdot\left(\frac{\sqrt{xz}}{y} \cdot \frac{\sqrt{yz}}{x} \cdot\frac{\sqrt{xy}}{z}\right)^{\frac13} \\ &=6\cdot(1)^{\frac13}=6 \end{align*}

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    $\begingroup$ Is this what you have in mind? $\endgroup$ – Lucian Apr 9 '14 at 17:06
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    $\begingroup$ You have $\frac{x}y$ plus it's reciprocal and three such sums. Each is at least $2$ by AM-GM. $\endgroup$ – Macavity Apr 9 '14 at 17:25
  • $\begingroup$ @Lucian and Macavity: Yes, once you see that the expression can be expressed as three reciprocals, the inequality is immediate. thanks. $\endgroup$ – user103828 Apr 9 '14 at 17:35
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$$ \frac{x+z}{y}+\frac{y+z}{x}+\frac{x+y}{z}\geq 6$$ is equivalent to $$ \frac{x^2 z+z^2x + y^2 z+z^2 y + x^2 y + y^2 x}{6}\geq xyz$$ which is true by the AM-GM inequality: $$ \frac{x^2 z+z^2x + y^2 z+z^2 y + x^2 y + y^2 x}{6}\geq \sqrt[6]{x^6y^6z^6}.$$

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  • $\begingroup$ yes, this is indeed a quick way to see it... as Lucian and Macavity mentioned in their comments another way to see it is if you see that the expression is three reciprocals. $\endgroup$ – user103828 Apr 9 '14 at 17:36

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