# A n-variable fractional inequality

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 $$0.
But, $$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.
• Thank you for your neat solution!(My friend and I didn't notice the Descartes' rule lol) Yes, the following statement ends the problem: WLOG, we assume $a_2=a_3=\dots=a_n=a, a_1=n-(n-1)a$ where $0< a \le \frac{n}{n-1}$ and $n\ge 3$, thus we need to prove $\frac{a^{n-1}(n-(n-1)a)}{(n-a)^{n-1}(n-1)a} \le \frac{(n-a)^{n-1}(n-1)a}{(n^2-2n+a)^{n-1}(n^2-n-(n-1)a)}$. When $a=1$ the equality holds. Otherwise, it is equivalent to $a^{n-3}(\frac{n}{n-1}-a)(n^2-2n+a)^{n-1} \le (n-a)^{2n-3}$. It's clear that LHS $\le 0 \le$ RHS. The equality holds iff $a=1$. So we proved what we want. Commented Oct 14, 2021 at 12:18
• A silly mistake of mine. LHS does not $\le 0$, but wo can still prove it by using derivative after taking logarithm :D Commented Oct 16, 2021 at 0:13