<s>[INST] 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...
 [/INST] 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 $b\le d$, $$\begin{align*}
\dfrac1{2\sqrt b} &\ge \dfrac1{2\sqrt d}\\
\sqrt c - \sqrt{c-1} &\ge \sqrt {c+1} - \sqrt{c}
\end{align*}$$
</s>