# 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.

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.