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.
 A: 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 $\displaystyle \frac{4}{ab} \ge \frac{2}{ac} \iff 2c \ge a \iff z \ge \frac{1}{2}$.
A: $f(x,y,z) = 2xy+2yz+2zx -x-y-z-3$ 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 $f(x,y,z) \geq 0$.
A: 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 $\frac{1}{2}$.


*

*$y=x$, $z=\frac{1}{x^2}$, 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!
