Prove that: $\sum\limits_{cyc}\frac{1}{a}\sum\limits_{cyc}\frac{1}{1+a^2}\geq\frac{16}{1+abcd}$ Let $a$, $b$, $c$ and $d$ be positive numbers. Prove that:
$$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\left(\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}+\frac{1}{1+d^2}\right)\geq\frac{16}{1+abcd}$$
I tried Rearrangement, C-S and more, but without success. 
 A: My second solution
By Cauchy-Bunyakovsky-Schwarz inequality, we have
\begin{align}
&\frac{1}{1+a^2} + \frac{1}{1+b^2} + \frac{1}{1+c^2} + \frac{1}{1+d^2}\\
=\ & \frac{(1/a)^2}{1+(1/a)^2} + \frac{(1/b)^2}{1+(1/b)^2} + \frac{(1/c)^2}{1+(1/c)^2} + \frac{(1/d)^2}{1+(1/d)^2}\\
\ge\ & \frac{(1/a + 1/b + 1/c + 1/d)^2}{4 + (1/a)^2 + (1/b)^2 + (1/c)^2 + (1/d)^2}.
\end{align}
It suffices to prove that
$$\frac{(1/a + 1/b + 1/c + 1/d)^3}{4 + (1/a)^2 + (1/b)^2 + (1/c)^2 + (1/d)^2}\ge \frac{16}{1 + abcd}$$
or equivalently
$$\frac{(a+b+c+d)^3}{4 + a^2 + b^2 + c^2 + d^2} \ge \frac{16abcd}{1 + abcd}.\tag{1}$$
By Vasc's Equal Variable Theorem [1, Corollary 1.9], we only need to prove the case when
$a=b=c \le d$. Let $d = a + s$ for $s \ge 0$. It suffices to prove that
\begin{align}
&a^3 s^4+(13 a^4-16 a^3+1) s^3+(60 a^5-48 a^4+12 a) s^2+(112 a^6-96 a^5-64 a^3+48 a^2) s\\
&\quad +64 a^7-64 a^6-64 a^4+64 a^3 \ge 0.
\end{align}
It is not difficult.
Remarks: I hope to see nice proofs for (1).
Reference
[1] Vasile Cirtoaje, “The Equal Variable Method”, J. Inequal. Pure and Appl. Math., 8(1), 2007.
% https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf
