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...

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.
