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.
 A: From the guidance of kingW3 I give a complete solution. We will prove that
$$
\sqrt{7}+\sqrt{11}<6, \quad \sqrt{32}+\sqrt{40}<12
$$
by taking squaring both sides of inequalities. Indeed, we have
$$
\sqrt{7}+\sqrt{11}<6\Leftrightarrow 7+11+2\sqrt{77}<36\Leftrightarrow\sqrt{77}<9,
$$
$$
\sqrt{32}+\sqrt{40}<12\Leftrightarrow 32+40+2\sqrt{1280}<144\Leftrightarrow\sqrt{1280}<36.
$$
A: $$\sqrt{7}+\sqrt{11}+\sqrt{32}+\sqrt{40}=\sqrt{7+40+2\sqrt{7\cdot40}}+\sqrt{11+32+2\sqrt{11\cdot32}}=$$
$$=\sqrt{47+2\sqrt{280}}+\sqrt{43+2\sqrt{352}}<\sqrt{47+2\cdot17}+\sqrt{43+2\cdot19}=18$$
A: $\sqrt{7}$ 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 $18-\left(\sqrt{7}+\sqrt{11}+\sqrt{32}+\sqrt{40}\right)\geq \frac{1}{18}$.
A: $$\begin{align}
3\left(\sqrt{7}+\sqrt{11}+\sqrt{32}+\sqrt{40}\right) 
&= \sqrt{63}+\sqrt{99}+\sqrt{288}+\sqrt{360}\\ 
&< \sqrt{64}+\sqrt{100}+\sqrt{289}+\sqrt{361}\\ 
&< 8+10+17+19\\
&< 3\times 18\end{align}$$
We can even estimate from $\sqrt{1+x}$ expansion that the error is about 
$\epsilon=-\frac 16(\frac 18+\frac 1{10}+\frac 1{17}+\frac 1{19})\approx -0.056\quad$ giving $\quad \approx 17.944$
A: Lets be methodical about this. Consider finding a nice upper bound for 
$\sqrt a + \sqrt b$.
$$(\sqrt a + \sqrt b)^2 = a + b + 2\sqrt{ab}$$
So maybe we should replace $ab$ with the smallest $n$ such that $ab \le n^2$
Then $(\sqrt a + \sqrt b)^2 < a + b + 2n$. So 
$$\sqrt a + \sqrt b < \sqrt{a + b + 2n}$$

We note that $7 \times 32 = 224 < 225 = 15^2$ and 
$11 \times 40 = 440 < 441 = 21^2$
Then
$$\sqrt 7 + \sqrt{32} < \sqrt{7 + 32 + 2\times 15} = \sqrt{69}$$
And
$$\sqrt{11} + \sqrt{40} < \sqrt{11 + 40 + 2\times 21} = \sqrt{93}$$
So 
$$\sqrt{69} + \sqrt{93} < \sqrt{69 + 93 + 2 \times 81} = 18$$
Hence $$\sqrt{7}+\sqrt{11}+\sqrt{32}+\sqrt{40} < 18$$
