<s>[INST] Compare sum of radicals I am stuck in a difficult question:
Compare $18$ and
$$
A=\sqrt{7}+\sqrt{11}+\sqrt{32}+\sqrt{40}
$$
without using calculator.
Thank you for all solution.
 [/INST] $\sqrt{7}$ and $\sqrt{11}$ can be written as $\sqrt{9\pm 2}$ and similarly $\sqrt{32}$ and $\sqrt{40}$ can be written as $\sqrt{36\pm 4}$.
Since $\sqrt{x}$ is a concave function on $\mathbb{R}^+$,
$$ \sqrt{9-2}+\sqrt{9+2}+\sqrt{36-4}+\sqrt{36+4} \color{red}{\leq} 2\sqrt{9}+2\sqrt{36} = 18.$$
We may also estimate the difference between the RHS and the LHS:
$$\begin{eqnarray*} 2n-\sqrt{n^2+x}-\sqrt{n^2-x}&=&\frac{x}{n+\sqrt{n^2-x}}-\frac{x}{n+\sqrt{n^2+x}}\\&=&x\cdot \frac{\sqrt{n^2+x}-\sqrt{n^2-x}}{(n+\sqrt{n^2-x})(n+\sqrt{n^2+x})}\\&=&\frac{2x^2}{(n+\sqrt{n^2-x})(n+\sqrt{n^2+x})(\sqrt{n^2+x}+\sqrt{n^2-x})}\\&\geq&\frac{x^2}{n(3n^2+\sqrt{n^4-x^2})}\geq\frac{x^2}{n\left(4n^2-\frac{x^2}{2n^2}\right)}.\end{eqnarray*} $$
This leads to $18-\left(\sqrt{7}+\sqrt{11}+\sqrt{32}+\sqrt{40}\right)\geq \frac{1}{18}$.
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