Iranian Mathematical Olympiad

$$\sum \left( \sqrt{\frac{a^2+1}{2}} - \sqrt{a} \right) = \sum \frac{\frac{a^2+1}{2} - a}{\sqrt{\frac{a^2+1}{2}} + \sqrt{a}} = \frac{1}{2} \sum \frac{(a-1)^2}{\sqrt{\frac{a^2+1}{2}} + \sqrt{a}}$$
According to Cauchy-Schwarz inequality, $\sqrt{\frac{a^2+1}{2}} + \sqrt{a} \le \sqrt{2(\frac{a^2+1}{2} + a)} = a + 1$, hence
$$\frac{1}{2} \sum \frac{(a-1)^2}{\sqrt{\frac{a^2+1}{2}} + \sqrt{a}} \ge \frac{1}{2} \sum \frac{(a-1)^2}{a+1}.$$
On the other hand, $\frac{(a-1)^2}{a+1} = (a-3) + \frac{4}{a+1}$; therefore,
$$\frac{1}{2} \sum \frac{(a-1)^2}{a+1} = \frac{1}{2} \sum (a-3) + 2 \underbrace{\sum \frac{1}{a+1}}_{2} = \frac{1}{2} (\sum a) - 2.$$
Now, it suffices to prove that $\frac{1}{2}(\sum a) - 2 \ge 2(\sum \sqrt{a}) - 8$, or equivalently, $3 + \frac{1}{4} \sum a \ge \sqrt{a}$, which can be deduced by the following AM-GM inequality
$$3 + \frac{1}{4} \sum a = \underbrace{\sum \frac{a}{a+1}}_{2} + \sum \frac{a+1}{4} = \sum \left( \frac{a}{a+1} + \frac{a+1}{4} \right) \ge \sum \sqrt{a}.$$