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$a_ile b_i,a_ile A,i=1,2,cdots,n$ 且$prodlimits_{i=1}^n{dfrac{b_i}{a_i}}ledfrac{B}{A}$.
证明:$prodlimits_{i=1}^n{dfrac{b_i+1}{a_i+1}}ledfrac{B+1}{A+1}$(2018全国联赛加试题第一题)
记$dfrac{b_i}{a_i}=1+x_i,x_ige0,(i=1,2,cdots)$
记$f_k=sumlimits_{1le i_1<i_2cdots<i_kle n}{x_{i_1}x_{i_2}cdots x_{i_k}}ge0$
则$prodlimits_{i=1}^{n}dfrac{1+b_i}{1+a_i}=prodlimits_{i=1}^n{dfrac{1+a_i(1+x_i)}{1+a_i}}le prodlimits_{i=1}^{n}dfrac{1+A(1+x_i)}{1+A}=prodlimits_{i=1}^{n}left(1+dfrac{A}{1+A}x_i
ight)$
$overset{ extbf{此处用到韦达定理}}{=}1+dfrac{A}{1+A}f_1+left(dfrac{A}{1+A}
ight)^2f_2+cdots+left(dfrac{A}{1+A}
ight)^nf_n$
$overset{ extbf{变形}}{=}dfrac{1+A(1+f_1+f_2+cdots+f_n)}{1+A}+dfrac{A}{1+A}sumlimits_{k=1}^nleft((dfrac{A}{1+A})^{k-1}-1)f_k
ight)$
$overset{ extbf{此处用到韦达定理}}{=}dfrac{1+Aprodlimits_{k=1}^n(1+x_i)}{1+A}+dfrac{A}{1+A}sumlimits_{k=1}^nleft((dfrac{A}{1+A})^{k-1}-1)f_k
ight)$
$ledfrac{1+Aprodlimits_{k=1}^n(1+x_i)}{1+A}ledfrac{1+B}{1+A}$