Problem: For positive numbers $a_1,a_2,\dots,a_n,$ note that $A=\sum\limits_{i=1}^{n}a_i, \,b_i=A-a_i,\,B=\sum\limits_{i=1}^{n}b_i.$ Prove $$ \frac{\prod\limits_{i=1}^{n}a_i}{\prod\limits_{i=1}^{n}(A-a_i)} \leqslant \frac{\prod\limits_{i=1}^{n}b_i}{\prod\limits_{i=1}^{n}(B-b_i)} $$ I know this inequality from a friend of mine who claimed the problem came from the Internet.
My friend and I tried it for days. It's clear that for $a_1=a_2=\dots=a_n=1$ the equality holds. The first thought is to use Jesen inequality after taking the logarithm. However for $f(x)=\ln\frac{x}{A-x}(A>x)$ we have $f''(x)=-\frac{A(A-2x)}{x^2(A-x)^2}$ which implies zero or one inflection point, so we can't use Jesen inequality directly.
Rewriting the inequality as $$ \prod\limits_{i=1}^{n}(B-b_i) \leqslant \frac{\prod\limits_{i=1}^{n}(A-a_i)^2}{\prod\limits_{i=1}^{n}a_i} $$ makes the fact that $\prod\limits_{i=1}^{n}(B-b_i)$ is hard to deal clear.
I wonder if there would be a nice solution.
Since our inequality is homogeneous, we can assume that $\sum\limits_{i=1}^na_i=n$.
Thus, $b_i=n-a_i,$ $\sum\limits_{i=1}^nb_i=n^2-n$ and we need to prove that: $$\frac{\prod\limits_{i=1}^na_i}{\prod\limits_{i=1}^n(n-a_i)}\leq\frac{\prod\limits_{i=1}^n(n-a_i)}{\prod\limits_{i=1}^n(n^2-n-b_i)}$$ or $$\frac{\prod\limits_{i=1}^na_i}{\prod\limits_{i=1}^n(n-a_i)}\leq\frac{\prod\limits_{i=1}^n(n-a_i)}{\prod\limits_{i=1}^n(n^2-2n+a_i)}$$ or $\sum\limits_{i=1}^nf(a_i)\geq0,$ where $$f(x)=2\ln(n-x)-\ln{x}-\ln(n^2-2n+x)$$ and $0<x<n$.
But, $$f''(x)=\frac{n^2(-2x^3-(n^2-6)x^2-2n(n-2)(n-3)x+n^2(n-2)^2)}{(n-x)^2(n^2-2n+x)^2x^2}$$ and by the Descartes' rule of signs (https://en.wikipedia.org/wiki/Descartes%27_rule_of_signs ) we see that $f''$ has on $(0,n)$ an unique root for any $n\geq3$.
Thus, by the Vasc's HCF Theorem it's enough to prove our inequality for equality case of $n-1$ variables.
Can you end it now?