Comparing the size of square roots How to compare the size of following numbers without using the calculator?
$a=\sqrt{2}+\sqrt{6}+\sqrt{7},$
$b=\sqrt{3}+\sqrt{4}+\sqrt{8},$
$c=\sqrt{5}+\sqrt{5}+\sqrt{5}$

Looking at the three expressions, one easily notices that the sum of the radicants is $15$ in all three cases. This together with the concavity of the square root ($\sqrt{\lambda a+(1-\lambda)b}\ge\lambda\sqrt{a}+(1-\lambda)\sqrt{b}$ for all $0\le\lambda\le1$) helps to compare the numbers. Note that the concavity of the square root is strict, so in the above equation, if $a\neq b$ and $\lambda\notin\{0,1\}$, the left hand side is strictly greater.
Clearly $c$ is the largest of these numbers, because here all three radicants are equal. So what remains is to compare $a$ with $b$.
It is also obvious that for a and b, additionally the product of the first two terms is equal. Therefore it is useful to square the numbers:
$$\begin{aligned}
a^2 &= 2 + 6 + 7 + 2 \sqrt{12} + 2 \sqrt{14} + 2 \sqrt{42}\\
b^2 &= 3 + 4 + 8 + 2 \sqrt{12} + 2 \sqrt{24} + 2 \sqrt{32}
\end{aligned}$$
Obviously the leading terms are equal, so ultimately we only have to compare $a' := 2 \sqrt{14} + 2 \sqrt{42} = \sqrt{8}(\sqrt{7} + \sqrt{21})$ with $b' := 2\sqrt{24}+2\sqrt{32} = \sqrt{8}(\sqrt{12}+\sqrt{16})$. Here we see that again, the sum of the radicants is the same, but for $a'$ their difference is larger, and therefore due to concavity $a'<b'$.
So putting everything together, we get
$$a < b < c.$$
