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.
The Buffalo Way works.
After clearing the denominators, it suffices to prove that $f(a,b,c,d)\ge 0$ where $f(a,b,c,d)$ is a polynomial. WLOG, assume that $a\le b\le c\le d$.
(1) If $1 \le a$, let $a = 1+s, \ b = 1+s+t, \ c = 1+s+t+r, \ d = 1+s+t+r+u; \ s, t, r, u\ge 0$. $f(1+s, 1+s+t, 1+s+t+r, 1+s+t+r+u)$ is a polynomial in $s, t, r, u$ with non-negative coefficients. True.
(2) If $a < 1 \le b \le c \le d$, let $a = \frac{1}{1+s}, \ b = 1+t, \ c = 1+t+r, \ d = 1+t+r+u; \ s, t, r, u\ge 0$. We have \begin{align} &(1+s)^4f(\frac{1}{1+s}, 1+t, 1+t+r, 1+t+r+u)\\ =\ & g_1(s, t, r, u) + 128r^2-64rs+320rt+128ru+48s^2-96st\\ &\quad -32su+240t^2+160tu+48u^2 \end{align} where $g_1(s,t,r,u)$ is a polynomial with non-negative coefficients. It suffices to prove that $$128r^2-64rs+320rt+128ru+48s^2-96st-32su+240t^2+160tu+48u^2 \ge 0.$$ True. However, hope to see a nice proof of it.
(3) If $a \le b < 1 \le c \le d$, let $a = \frac{1}{1+s+t}, \ b = \frac{1}{1+s}, \ c = 1+r, \ d = 1+r+u; \ s, t, r, u\ge 0$. We have \begin{align} &(1+s+t)^4(1+s)^4f(\frac{1}{1+s+t}, \frac{1}{1+s}, 1+r, 1+r+u) \\ =\ & g_2(s,t,r,u) + 128r^2-128rs-64rt+128ru+128s^2+128st \\ &\quad -64su+48t^2-32tu+48u^2 \end{align} where $g_2(s,t,r,u)$ is a polynomial with non-negative coefficients. It suffices to prove that $$128r^2-128rs-64rt+128ru+128s^2+128st-64su+48t^2-32tu+48u^2\ge 0.$$ True. However, hope to see a nice proof of it.
(4) If $a\le b \le c < 1 \le d$, let $a = \frac{1}{1+s+t+r}, \ b = \frac{1}{1+s+t}, \ c = \frac{1}{1+s}, \ d = 1+u; \ s,t,r,u\ge 0$. We have \begin{align} &(1+s+t+r)^4(1+s+t)^4(1+s)^4f(\frac{1}{1+s+t+r}, \frac{1}{1+s+t}, \frac{1}{1+s}, 1+u)\\ =\ & g_3(s,t,r,u) + 32 r^3+384 r^2 s+240 r^2 t+48 r^2 u+1168 r s^2+1552 r s t \\ &\quad -64 r s u+528 r t^2+16 r t u+48 r u^2+1168 s^3+2336 s^2 t-96 s^2 u\\ &\quad +1552 s t^2-128 s t u+144 s u^2+352 t^3+16 t^2 u+96 t u^2+64 u^3\\ &\quad +48 r^2+160 r s+128 r t-32 r u+240 s^2+320 s t-96 s u+128 t^2-64 t u+48 u^2 \end{align} where $g_3(s,t,r,u)$ is a polynomial with non-negative coefficients. It suffices to prove that \begin{align} &32 r^3+384 r^2 s+240 r^2 t+48 r^2 u+1168 r s^2+1552 r s t \\ &\quad -64 r s u+528 r t^2+16 r t u+48 r u^2+1168 s^3+2336 s^2 t-96 s^2 u\\ &\quad +1552 s t^2-128 s t u+144 s u^2+352 t^3+16 t^2 u+96 t u^2+64 u^3\\ &\quad +48 r^2+160 r s+128 r t-32 r u+240 s^2+320 s t-96 s u+128 t^2-64 t u+48 u^2\ge 0. \end{align} True. However, hope to see a nice proof of it.
(5) If $a\le b\le c\le d < 1$, let $a = \frac{1}{1+s+t+r+u}, \ b = \frac{1}{1+s+t+r}, \ c = \frac{1}{1+s+t}, \ d = \frac{1}{1+s}; \ s,t,r,u\ge 0$. $(1+s+t+r+u)^4(1+s+t+r)^4(1+s+t)^4(1+s)^4f(\frac{1}{1+s+t+r+u}, \frac{1}{1+s+t+r}, \frac{1}{1+s+t}, \frac{1}{s})$ is a polynomial in $s, t, r, u$ with non-negative coefficients. True.
We are done.
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
My third proof.
By Cauchy-Bunyakovsky-Schwarz inequality, we have $$\mathrm{LHS} \ge \left(\sum_{\mathrm{cyc}} \frac{1}{\sqrt{a(1+a^2)}}\right)^2.$$
It suffices to prove that $$\sum_{\mathrm{cyc}} \frac{1}{\sqrt{a(1+a^2)}} \ge \frac{4}{\sqrt{1 + abcd}}.$$
Using $2(1 + abcd) \ge (1 + \sqrt{abcd})^2$, it suffices to prove that $$\sum_{\mathrm{cyc}} \frac{1}{\sqrt{a(1+a^2)}} \ge \frac{4\sqrt 2}{1 + \sqrt{abcd}}$$ or $$\sum_{\mathrm{cyc}} \frac{1}{\sqrt{2a(1+a^2)}} \ge \frac{4}{1 + \sqrt{abcd}}. \tag{1}$$
Let $a = \mathrm{e}^x, b = \mathrm{e}^y, c = \mathrm{e}^z, d = \mathrm{e}^w$. We need to prove that $$\sum_{\mathrm{cyc}} \frac{1}{\sqrt{2\mathrm{e}^x(1+\mathrm{e}^{2x})}} \ge \frac{4}{1 + \mathrm{e}^{(x+y+z+w)/2}}. \tag{2}$$
Let $$f(x) := \frac{1}{\sqrt{2\mathrm{e}^x(1+\mathrm{e}^{2x})}}.$$ We have $$f''(x) = \frac{\sqrt 2\, (9\mathrm{e}^{4x} - 2\mathrm{e}^{2x} + 1)}{8(1+\mathrm{e}^{2x})^2\sqrt{\mathrm{e}^x(1+\mathrm{e}^{2x})}} > 0.$$ Thus, $f(x)$ is convex on $\mathbb{R}$.
By Jensen and AM-GM, we have $$\mathrm{LHS}_{(2)} \ge 4 f\left(\frac{x + y + z + w}{4}\right) = \frac{4}{\sqrt{2q(1+q^2)}} \ge \frac{4}{1 + q^2} = \mathrm{RHS}_{(2)}$$ where $q = \mathrm{e}^{(x+y+z+w)/4}$.
We are done.