Q:

Inequality under condition $a+b+c=0$

I don't know how to prove that the following inequality holds (under condition $a+b+c=0$):
$$\frac{(2a+1)^2}{2a^2+1}+\frac{(2b+1)^2}{2b^2+1}+\frac{(2c+1)^2}{2c^2+1}\geqq 3$$

A:

This one made me struggle so much that I was close to go crazy. Therefore let me post my solution to have a relief from this burden..
First we have $$2a^2=\frac43a^2+\frac23a^2=\frac43a^2+\frac23(b+c)^2\leq \frac43(a^2+b^2+c^2);$$ where the last inequality follows from the arithmetic-quadratic mean.
Analogously $$\begin{split}2b^2&\leq \frac43 (a^2+b^2+c^2),\\ 2c^2&\leq \frac43(a^2+b^2+c^2).\end{split}$$ It follows that $$\sum_\text{cyc}\frac{(2a+1)^2}{2a^2+1}\geq3\sum_\text{cyc}\frac{(2a+1)^2}{4(a^2+b^2+c^2)+3}=3\left(1+\frac{4(a+b+c)}{4(a^2+b^2+c^2)+3}\right)=3.$$

