Proving $\sum\limits_{\text{cyc}} \frac{a}{b^2+c^2+d^2} \geq \frac{3\sqrt{3}}{2}\frac{1}{\sqrt{a^2+b^2+c^2+d^2}}$ for $a, b, c, d >0$ 
Let $a, b, c, d > 0$. Prove that
$$\frac{a}{b^2+c^2+d^2}+\frac{b}{a^2+c^2+d^2}+\frac{c}{a^2+b^2+d^2}+\frac{d}{a^2+b^2+c^2}$$
$$\geq\frac{3\sqrt{3}}{2}\frac{1}{\sqrt{a^2+b^2+c^2+d^2}}.$$

What I tried, was to say that $a^2+b^2+c^2+d^2=1$ and so
$$\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}+\frac{d}{1-d^2}≥\frac{3\sqrt{3}}{2}$$ and
$$-\frac{a}{(a+1)(a-1)}-\frac{b}{(b+1)(b-1)}-\frac{c}{(c+1)(c-1)}-\frac{d}{(d+1)(d-1)}≥\frac{3\sqrt{3}}{2}$$
and so
$$\frac{a}{a+1}-\frac{a}{a-1}+\frac{b}{b+1}-\frac{b}{b-1}+\frac{c}{c+1}-\frac{c}{c-1}+\frac{d}{d+1}-\frac{d}{d-1}≥\frac{3\sqrt{3}}{2}$$
Perhaps this leads somewhere?

Since the desired inequality is homogeneous, assume that $a^2 + b^2 + c^2 + d^2 = 1$. The desired inequality becomes
$$\sum_{\mathrm{cyc}}\frac{a}{1 - a^2} \ge \frac{3\sqrt 3}{2}.$$
Note that
\begin{align*}
 \frac{a}{1 - a^2} - \frac{3\sqrt 3}{2}a^2
 &= \frac{a}{2(1 - a^2)}
 [2 - 3\sqrt 3\, a(1 - a^2)]\\
 &= \frac{a}{2(1 - a^2)}\frac{4 - 27a^2(1 - a^2)^2}{2 + 3\sqrt 3\, a(1 - a^2)}\\
 &\ge 0
\end{align*}
where we have used AM-GM to get
$2a^2(1 - a^2)^2 \le \left(\frac{2a^2 + 1-a^2 + 1 - a^2}{3}\right)^3 = \frac{8}{27}$.
Thus, we have
$$\sum_{\mathrm{cyc}} \frac{a}{1 - a^2}
\ge \frac{3\sqrt 3}{2}(a^2 + b^2 + c^2 + d^2) = \frac{3\sqrt 3}{2}.$$
We are done.
