<s>[INST] 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.
 [/INST] $f(x,y,z) = 2xy+2yz+2zx -x-y-z-3$ subject to $xyz=1$.
$f_x = \lambda zy \to 2xy+2xz - x = \lambda$.
$f_y = \lambda xz \to 2xy + 2yz - y = \lambda$.
$\Rightarrow  (2z-1)(x-y) = 0$.
If $z=\dfrac{1}{2} \to f(x,y,z) = f(x,y,\frac{1}{2}) = 2xy + y+x-x-y-\dfrac{7}{2}=2(2)-\dfrac{7}{2} = \dfrac{1}{2}>0$
Thus assume $x=y$, and other cases gives: $x=y=z \to x =y=z = 1 \to f(1,1,1) = 0$. Thus $f(x,y,z) \geq 0$.
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