Sixteenth STARS OF MATHEMATICS Competition

Replace $(c, d)$ with $(-c, -d)$ to rephrase the problem as: subject to the constraint $ab(c+d) + cd(a+b) = 0$, prove that $\sum \frac{a+1}{a^2+3} \ge 0$ (1).

Suppose one of the numbers equals $0$ and notice that (at least) another is $0$ as well – let them be $c$ and $d$. The claim (1) rewrites as
$$\frac{a+1}{a^2+3} + \frac{b+1}{b^2+3} \ge -\frac{2}{3},$$
which holds true as $\frac{x+1}{x^2+3} \ge -\frac{1}{3}$, for all $x \in \mathbb{R}$.

Suppose now that $a$, $b$, $c$, $d \ne 0$ and set $x = \frac{1}{a}$, $y = \frac{1}{b}$, $z = \frac{1}{c}$ and $t = \frac{1}{d}$. Once again, the problem is restated:
$$x + y + z + t = 0 \implies \sum \frac{x^2 + x}{3x^2 + 1} \ge 0. \quad (2)$$
Let $|t| = \max\{|x|, |y|, |z|, |t|\}$. The inequality (2) is equivalent to
$$\sum \left( \frac{6x^2 + 6x}{3x^2 + 1} + 1 \right) \ge 4 \iff \sum \frac{(3x+1)^2}{3x^2 + 1} \ge 4$$
$$\iff \frac{(3x+1)^2}{3x^2+1} + \frac{(3y+1)^2}{3y^2+1} + \frac{(3z+1)^2}{3z^2+1} \ge 4 - \frac{(3t+1)^2}{3t^2+1} = \frac{3(1-t)^2}{3t^2+1}.$$
Notice that $x^2+y^2+z^2 \le 3t^2$ and apply Cauchy-Schwarz inequality to conclude:
$$\frac{(3x+1)^2}{3x^2+1} + \frac{(3y+1)^2}{3y^2+1} + \frac{(3z+1)^2}{3z^2+1} \ge \frac{3(x+y+z+1)^2}{x^2+y^2+z^2+1} \ge \frac{3(1-t)^2}{3t^2+1}.$$