Prove that $\sqrt{ c} − \sqrt{c − 1} \geq \sqrt{ c + 1} −\sqrt{c}$ for all real $c \geq 1$. Prove that $\sqrt{ c} − \sqrt{c − 1} \geq \sqrt{c + 1} −\sqrt{c}$ for all real $c \geq 1$.
Can anyone provide some form of guidance? So far all I have been able to think of is writing $c$ as $x^2$ for some $x$, or eliminating the radical on one side...

To show
$\sqrt{ c} − \sqrt{c − 1} 
\geq \sqrt{c + 1} −\sqrt{c}
$.
This is the same as
$\sqrt{c}
\ge \frac{\sqrt{c + 1}+\sqrt{c - 1}}{2}
$.
If 
$f(x) = x^{1/2}
$,
then
$f'(x) = \frac12x^{-1/2}
$
and
$f''(x) = -\frac14x^{-3/2}
< 0
$
so
$f(x)$
is concave
which means that
$f(\frac{a+b}{2})
\ge \frac{f(a)+f(b)}{2}
$.
