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.

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 $$\sqrt{7}+\sqrt{11}+\sqrt{32}+\sqrt{40} < 18$$
