Equivalence of two inequalities

Question

If $a,b,c$ are positive real numbers and $f(x) = \dfrac{1}{\sqrt{1+e^x}}$, why are these two inequalities equivalent? $$\frac{1}{\sqrt{1+(abc)^{1/3}}} \le \frac{1}{3}\left( \frac{1}{\sqrt{1+a}} + \frac{1}{\sqrt{1+b}} + \frac{1}{\sqrt{1+c}} \right),\\f\left( \frac{x+y+z}{3}\right) \leq \frac{1}{3}\left(f(x)+f(y)+f(z)\right)$$

Answer 1

Just put $a=e^{x},b=e^{y},c=e^{z}$.