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.

We make the usual substitution $x = \dfrac{b}{a},y = \dfrac{c}{b}$ and $z = \dfrac{a}{c}$, then the inequality translates into proving:
$$\displaystyle 2\left(\sum\limits_{cyc} \frac{a}{b} - 3\right) \ge \sum\limits_{cyc} \frac{b}{a} - 3$$
Wlog assume $c$ lies between $a$ and $b$ (that is $a \le c \le b$ or $b \le c \le a$)
We have $\displaystyle \sum\limits_{cyc} \frac{a}{b} - 3 = \frac{(a-b)^2}{ab} - \frac{(a-c)(c-b)}{ac}$
and $\displaystyle \sum\limits_{cyc} \frac{b}{a} - 3 = \frac{(a-b)^2}{ab} - \frac{(a-c)(c-b)}{bc}$
Thus we are required to show: $\displaystyle \frac{(a-b)^2}{ab} \ge (a-c)(c-b)\left(\frac{2}{ac} - \frac{1}{bc}\right)$
Since, $(a-b)^2 = (a-c + c-b)^2 \ge 4(a-c)(c-b) \ge 0$ it suffices to prove 
$\displaystyle \frac{4}{ab}+\frac{1}{bc} \ge \frac{2}{ac}$ which is true since $\displaystyle \frac{4}{ab} \ge \frac{2}{ac} \iff 2c \ge a \iff z \ge \frac{1}{2}$.
