<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] 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.
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