2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.