If $x,y,z \geq 1/2, xyz=1$, showing that $2(1/x+1/y+1/z) \geq 3+x+y+z$ If $x,y,z \geq 1/2, xyz=1$, showing that $2(1/x+1/y+1/z) \geq 3+x+y+z$
I tried Schturm's method for quite some time, and Cauchy Schwarz for numerators because of the given product condition.

Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, our inequality is equivalent to $f(v^2)\geq0$, where $f$ is a linear increasing function.
Hence, $f$ gets a minimal value, when $v^2$ gets a minimal value, which happens for equality of two variables or maybe one of them equal to $\frac{1}{2}$.


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*$y=x$, $z=\frac{1}{x^2}$, which gives $(x-1)^2(2x^2+2x-1)\geq0$;

*$z=\frac{1}{2}$, $y=\frac{2}{x}$, which gives $\frac{1}{2}\geq0$. Done!
