AM-GM inequality

Question

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*}

Answer 1

$ \frac{x+y}{z} +\frac{y+z}{x} +\frac{z+x}{y} = \frac{(x+y)^2}{(x+y)z} +\frac{(y+z)^2}{(y+z)x} +\frac{(z+x)^2}{(z+x)y} \overset{Bergstrom}{\geqslant} \dfrac{(2(x+y+z))^2}{2(xy+yz+zx)} = $

$= 2 \cdot \frac{(x+y+z)^2 }{xy+yz+zx}$

Since $(x+y+z)^2 \geqslant 3(xy+yz+zx) $

then $ 2 \cdot \frac{(x+y+z)^2 }{xy+yz+zx} \geqslant 2\cdot 3$

Answer 2

$$ \frac{x+z}{y}+\frac{y+z}{x}+\frac{x+y}{z}\geq 6$$ 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: $$ \frac{x^2 z+z^2x + y^2 z+z^2 y + x^2 y + y^2 x}{6}\geq \sqrt[6]{x^6y^6z^6}.$$