$$\sum\frac{a+b}{a^2+b^2}\leq\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{abc}}=\sum\frac{1}{\sqrt{ab}}$$The above inequality follows from $$\frac{a+b}{a^2+b^2}\leq\frac{1}{\sqrt{ab}}$$$$\iff a^2+b^2\geq\sqrt{a^3b}+\sqrt{ab^3}$$which can be easily proven by Muirhead, rearrangement or AM-GM, for example $$\frac{3a^2+b^2}{4}+\frac{a^2+3b^2}{4}\geq\sqrt{a^3b}+\sqrt{ab^3}$$

L3435 wrote: $$\frac{a+b}{a^2+b^2}\leq\frac{1}{\sqrt{ab}}$$$$\iff a^2+b^2\geq\sqrt{a^3b}+\sqrt{ab^3}$$which can be easily proven by Muirhead, rearrangement or AM-GM, for example $$\frac{3a^2+b^2}{4}+\frac{a^2+3b^2}{4}\geq\sqrt{a^3b}+\sqrt{ab^3}$$ you can prove by: $\frac{a+b}{a^2+b^2}\leq \frac{2(a+b)}{(a+b)^2}\leq \frac{1}{\sqrt{ab}}$

This inequality could be proven if this part is true: $\frac{a+b}{a^2+b^2} \leq \frac{\sqrt{c}}{\sqrt{2015}} = \frac{\sqrt{c}}{\sqrt{abc}} = \frac{1}{\sqrt{ab}}$ Then: $\frac{a^2+b^2}{a+b} \geq \sqrt{ab}$ $(a^2+b^2)^2 \geq ab(a+b)^2$ $a^4 + 2(ab)^2 + b^4 \geq ab(a^2 + 2ab + b^2) = a^3b + 2(ab)^2 + ab^3$ $a^4 - a^3b + b^4 - ab^3 \geq 0$ $(a-b)(a^3-b^3) \geq 0$ And this is trivially true.