MT【216】韦达定理

设$n$为正整数,$a_1,a_2,\cdots,a_n;b_1,b_2,\cdots,b_n;A,B$都是正数,
满足$a_i\le b_i,a_i\le A,i=1,2,\cdots,n$ 且$\prod\limits_{i=1}^n{\dfrac{b_i}{a_i}}\le\dfrac{B}{A}$.
证明:$\prod\limits_{i=1}^n{\dfrac{b_i+1}{a_i+1}}\le\dfrac{B+1}{A+1}$(2018全国联赛加试题第一题)

记$\dfrac{b_i}{a_i}=1+x_i,x_i\ge0,(i=1,2,\cdots)$
记$f_k=\sum\limits_{1\le i_1<i_2\cdots<i_k\le n}{x_{i_1}x_{i_2}\cdots x_{i_k}}\ge0$
则$\prod\limits_{i=1}^{n}\dfrac{1+b_i}{1+a_i}=\prod\limits_{i=1}^n{\dfrac{1+a_i(1+x_i)}{1+a_i}}\le \prod\limits_{i=1}^{n}\dfrac{1+A(1+x_i)}{1+A}=\prod\limits_{i=1}^{n}\left(1+\dfrac{A}{1+A}x_i\right)$
$\overset{\textbf{此处用到韦达定理}}{=}1+\dfrac{A}{1+A}f_1+\left(\dfrac{A}{1+A}\right)^2f_2+\cdots+\left(\dfrac{A}{1+A}\right)^nf_n$
$\overset{\textbf{变形}}{=}\dfrac{1+A(1+f_1+f_2+\cdots+f_n)}{1+A}+\dfrac{A}{1+A}\sum\limits_{k=1}^n\left((\dfrac{A}{1+A})^{k-1}-1)f_k\right)$
$\overset{\textbf{此处用到韦达定理}}{=}\dfrac{1+A\prod\limits_{k=1}^n(1+x_i)}{1+A}+\dfrac{A}{1+A}\sum\limits_{k=1}^n\left((\dfrac{A}{1+A})^{k-1}-1)f_k\right)$
$\le\dfrac{1+A\prod\limits_{k=1}^n(1+x_i)}{1+A}\le\dfrac{1+B}{1+A}$