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

The following proof, as it is, only works for $a,b,c \not\in (-2,0)$.
Let $|a| \geq |b| \geq |c|$
$$
\frac{(2x+1)^2}{2x^2+1} = 1 + \frac{2x^2+4x}{2x^2+1}
$$
Then it follows that
$$
\text{left-hand side}
= 3 + \frac{2a^2+4a}{2a^2+1} + \frac{2b^2+4b}{2b^2+1} + \frac{2c^2+4c}{2c^2+1}
\geq 3 + \frac{2(a^2+b^2+c^2)+4(a+b+c)}{2a^2+1}
=
$$
$$
= 3 + \frac{2(a^2+b^2+c^2)}{2a^2+1}
\geq 3
$$
Now in the second inequality I assumed for $x\in\{a,b,c\}$ that $0 \leq 2x^2+4x = 2(x^2+2x) = 2x(x+2)$. As the parabola defined by $2x(x+2)$ only is negative between -2 and 0 exclusive we can do so savely if none of $a,b,c$ are in the interval $(-2,0)$.
As $2a^2+1\geq 2b^2+1$ and $2a^2 +1 \geq 2c^2+1$ we get by multiplying with $\frac{2b^2+1}{2a^2+1} \leq 1$ and $\frac {2c^2+1}{2a^2+1} \leq 1$ that
$$\frac{2b^2+4b}{2b^2+1} \geq \frac{2b^2+4b}{2b^2+1}\cdot\frac{2b^2+1}{2a^2+1}=\frac{2b^2+4b}{2a^2+1}$$ etc.
Reducing to a common denominator and applying $a+b+c=0$ we get the inequality.
