Let $\displaystyle f(x) = \frac {e^x}{1 + e^x}$
a. Show that $\displaystyle f(x) > \frac {1}{2}$ for all $\displaystyle x >= 0$
$\displaystyle \frac {e^x}{1 + e^x} > \frac 12$
cross multiply
$\displaystyle 2e^x > 1 + e^x$
combine like terms
$\displaystyle e^x > 1$
take natural log
$\displaystyle x > 0$
Note that this is not what you posted, you said when x = 0, f(x) > 1/2. But we have that this is not true. It can be verified by plugging in 0 and seeing that your initial criteria are not correct:
$\displaystyle f(0) = \frac {e^0}{1 + e^0}$
$\displaystyle f(0) = \frac {1}{1 + 1}$
$\displaystyle f(0) = \frac 12$
and since 1/2 is not greater than 1/2, your initial criteria should have been either
a. Show that $\displaystyle f(x) \geq \frac {1}{2}$ for all $\displaystyle x \geq 0$
or
a. Show that $\displaystyle f(x) > \frac {1}{2}$ for all $\displaystyle x > 0$