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...
 A: We have:
$$
\sqrt{c}-\sqrt{c-1}≥\sqrt{c+1}-\sqrt{c}\iff\\
\\
\left(\sqrt{c}-\sqrt{c-1}\right)\frac{\sqrt{c}+\sqrt{c-1}}{\sqrt{c}+\sqrt{c-1}}≥\left(\sqrt{c+1}-\sqrt{c}\right)\frac{\sqrt{c+1}+\sqrt{c}}{\sqrt{c+1}+\sqrt{c}}\\
\\
\frac{1}{\sqrt{c}+\sqrt{c-1}}≥\frac{1}{\sqrt{c+1}+\sqrt{c}}\iff\\
\\
\sqrt{c+1}+\sqrt{c}≥\sqrt{c}+\sqrt{c-1}\iff\\
\\
\sqrt{c+1}≥\sqrt{c-1}
$$
and the last one is trivial.
A: 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}
$.
A: Hint:
study the variations of the function:
f(x) = (left hand side of your inequality)
this should then yield the answer very quickly
A: Note that $\sqrt{c+1}\geq \sqrt{c-1}$ and so $\frac{1}{\sqrt{c}+\sqrt{c-1}}\geq \frac{1}{\sqrt{c}+\sqrt{c+1}}$. Write $\sqrt{c}-\sqrt{c-1}=\frac{1}{\sqrt{c}+\sqrt{c-1}}$ and $\sqrt{c+1}-\sqrt{c}=\frac{1}{\sqrt{c}+\sqrt{c+1}}.$ Thus you have the result.
A: Consider $$B = \sqrt c - \sqrt{c-1},\quad D = \sqrt {c+1} - \sqrt{c}$$
By mean value theorem, there exist $$b\in (c-1, c),\quad d\in (c, c+1)$$ that satisfy
$$\frac1{2\sqrt b} = B,\quad \frac1{2\sqrt d} = D$$
and since $b\le d$, $$\begin{align*}
\dfrac1{2\sqrt b} &\ge \dfrac1{2\sqrt d}\\
\sqrt c - \sqrt{c-1} &\ge \sqrt {c+1} - \sqrt{c}
\end{align*}$$
A: Rewrite as: $$2\sqrt{ c} \geq \sqrt{ c + 1} +\sqrt{c-1}$$ and since both sides are nonegative it is equaivalent if we square it:
$$4c\geq c+1+2\sqrt{c^2-1}+c-1$$ or $$c\geq \sqrt{c^2-1}\iff c^2\geq c^2-1$$
which is true.
