This solution is almost surely not the best, but it works.
$\color{#FF6200}{\bf{Lemma.}}$ If $\lambda\in\mathbb{R}$, then $\forall X\geq4\lambda:Q_\lambda(X):=X^2-\lambda(2+\lambda)X+\lambda^2(1-\lambda)^2\geq0$.
Proof. Note that the discriminant is $\Delta=\lambda^2(2+\lambda)^2-4\lambda^2(1-\lambda)^2=3\lambda^3(4-\lambda)$. So, if $\lambda\not\in[0,4]$, then $\Delta<0$ which means that it has no real solutions and, since the polynomial is $\geq0$ at $X=0$, it follows that it is $\geq0$ everywhere (in particular for $X\geq4\lambda$). So let's study the case $\lambda\in[0,4]$. Note that
$$Q_\lambda(4\lambda)=16\lambda^2+\lambda(2+\lambda)4\lambda+\lambda^2(1-\lambda)^2=\lambda^4-6\lambda^3+9\lambda^2=\lambda^2(\lambda-3)^2\geq0$$
Moreover, $4\lambda\geq\lambda\cdot3\geq\frac{1}{2}\lambda(2+\lambda)$ where $X=\frac{1}{2}\lambda(2+\lambda)$ is the vertex of the convex parabola. But this means that $\forall X\geq 4\lambda$ the function $Q_\lambda(X)$ is increasing and, thus, $Q_\lambda(X)\geq Q_\lambda(4\lambda)\geq0$. $\ \square$
$\color{#FF3200}{\bf{Proposition.}}$ If $x+y+z=0$ are real numbers not in $\frac{\pi}{2}+\pi\mathbb{Z}$, then
$$\sum_\text{cyc}\tan^2(x)\tan^2(y)\geq\left(\sum_\text{cyc}\tan(z)\right)^2$$
Proof.
From now on let's denote $P:=\tan(x)\tan(y)$ and $S:=\tan(x)+\tan(y)$. Note that $$z=-(x+y)\Rightarrow\tan(z)=-\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}=-\frac{S}{1-P}$$
Now observe that, by the previous lemma, since $S^2\geq 4P$ by the AM-GM inequality
$$S^4-P(2+P)S^2+P^2(1-P)^2\geq0$$
$$\Rightarrow S^4-2PS^2+P^2(1-P)^2\geq P^2S^2$$
$$\Rightarrow S^2(S^2-2P)+P^2(1-P)^2\geq P^2S^2$$
Now, if $P=1$ we would then have
$$\tan(x)=\frac{1}{\tan(y)}=\tan\left(\frac{\pi}{2}-y\right)\Rightarrow z=-(x+y)\in\frac{\pi}{2}+\pi\mathbb{Z}$$
which contradicts the hypothesis. Hence, we can divide by $(1-P)^2$ on the previous inequality
$$\frac{S^2}{(1-P)^2}(S^2-2P)+P^2\geq \frac{P^2S^2}{(1-P)^2}$$
$$\Rightarrow P^2+(S^2-2P)\left(\frac{-S}{1-P}\right)^2\geq\left(S-\frac{S}{1-P}\right)^2$$
$$\Rightarrow \tan^2(x)\tan^2(y)+(\tan^2(x)+\tan^2(y))\tan^2(z)\geq\left(\tan(x)+\tan(y)+\tan(z)\right)^2$$
$$\therefore\sum_\text{cyc}\tan^2(x)\tan^2(y)\geq\left(\sum_\text{cyc}\tan(z)\right)^2$$
As an observation, note that we have to assume that neither of $x,y,z$ are in $\frac{\pi}{2}+\pi\mathbb{Z}$ because, otherwise, the tangent isn't defined. These degenerate cases come from the step of dividing by $|\prod\limits_\text{cyc}(a+b)|^2$ where you have to assume that it is $\neq0$. But this is not too big of a deal since the case $a+b=0$ is quite easy to handle (and the other degenerate cases follow directly from this one by symmetry).
