﻿ 一类条件代数不等式的注记

# 一类条件代数不等式的注记A Note on a Class of Conditional Algebraic Inequalities

Abstract: This paper mainly discusses a class of conditional algebraic inequalities. The main results are the supplements and extensions of the existing results and methods.

1. 引言

$\left(\frac{1}{{a}^{m}}-{a}^{n}\right)\left(\frac{1}{{b}^{m}}-{b}^{n}\right)\ge {\left(\frac{{2}^{m+n}-1}{{2}^{n}}\right)}^{2}$ (1)

$\underset{i=1}{\overset{n}{\prod }}\left(\frac{1}{{a}_{i}^{p}}-{a}_{i}^{q}\right)\ge {\left({n}^{p}-\frac{1}{{n}^{q}}\right)}^{n}$ (2)

$\underset{i=1}{\overset{n}{\prod }}\left(\frac{1}{{a}_{i}^{}}-{a}_{i}^{q}\right)\ge {\left({n}^{}-\frac{1}{{n}^{q}}\right)}^{n}$ (3)

(3)式可以看成是(2)式的补遗。(3)式中 $n=2$$q\ge 3$，一个自然的问题是如果 $n=2$$q=2$，(3)式是否成立。另外，文献 [7] 在证明(3)式 $n=2$$q\ge 3$ 的情形时证明技巧较高，也较复杂。本文将给出较直接与自然的证明。我们的结果可以看成是(1)与(3)式的补遗和扩展。我们的主要结论如下：

$\left(\frac{1}{a}-{a}^{2}\right)\left(\frac{1}{b}-{b}^{2}\right)\le {\left(\frac{7}{4}\right)}^{2}$ (4)

$f\left(x\right)=\left(\frac{1}{x}-{x}^{2}\right)\left(\frac{1}{1-x}-{\left(1-x\right)}^{2}\right)$.

$\begin{array}{c}f\left(x\right)=\frac{\left(1-{x}^{3}\right)\left(1-{\left(1-x\right)}^{3}\right)}{x\left(1-x\right)}=\left(1+x+{x}^{2}\right)\left(1+\left(1-x\right)+{\left(1-x\right)}^{2}\right)\\ ={x}^{4}-2{x}^{3}+{x}^{2}+3\end{array}$

$f\left(x\right)$ 求导得

${f}^{\prime }\left(x\right)=4{x}^{3}-6{x}^{2}+2x=2x\left(x-1\right)\left(2x-1\right)$

${f}^{\prime }\left(x\right)=0$，得到 ${f}^{\prime }\left(x\right)$ 在区间 $\left(0,1\right)$ 上的唯一零点 $x=\frac{1}{2}$，且

$0 时， ${f}^{\prime }\left(x\right)>0$ ；当 $\frac{1}{2} 时， ${f}^{\prime }\left(x\right)<0$

$f\left(x\right)$ 在区间 $\left(0,1\right)$ 上取到最大值

$f\left(\frac{1}{2}\right)={\left(2-\frac{1}{{2}^{2}}\right)}^{2}={\left(\frac{7}{4}\right)}^{2}$.

$\left(\frac{1}{a}-{a}^{p}\right)\left(\frac{1}{b}-{b}^{p}\right)\ge {\left(2-\frac{1}{{2}^{p}}\right)}^{2}$ (5)

${f}_{p}\left(x\right)=\left(\frac{1}{x}-{x}^{p}\right)\left(\frac{1}{1-x}-{\left(1-x\right)}^{p}\right)$, (6)

$\begin{array}{c}{f}_{p}\left(x\right)=\frac{\left(1-{x}^{p+1}\right)\left(1-{\left(1-x\right)}^{p+1}\right)}{x\left(1-x\right)}\\ =\left(1+x+\cdots +{x}^{p}\right)\left(1+\left(1-x\right)+\cdots +{\left(1-x\right)}^{p}\right).\end{array}$

${{f}^{\prime }}_{p}\left(x\right)=\left(1-2x\right){g}_{p}\left(x\right)$ (7)

$\begin{array}{c}{{f}^{\prime }}_{3}\left(x\right)=\left(1+2x+3{x}^{2}\right)\left(1+\left(1-x\right)+{\left(1-x\right)}^{2}+{\left(1-x\right)}^{3}\right)\\ -\left(1+x+{x}^{2}+{x}^{3}\right)\left(1+2\left(1-x\right)+3{\left(1-x\right)}^{2}\right)\\ =1+\left(1-x\right)+{\left(1-x\right)}^{2}+{\left(1-x\right)}^{3}-1-x-{x}^{2}-{x}^{3}\\ +2x+2x\left(1-x\right)+2x{\left(1-x\right)}^{2}+2x{\left(1-x\right)}^{3}\\ -2\left(1-x\right)-2x\left(1-x\right)-2{x}^{2}\left(1-x\right)-2{x}^{3}\left(1-x\right)\end{array}$

$\begin{array}{c}+3{x}^{2}+3{x}^{2}\left(1-x\right)+3{x}^{2}{\left(1-x\right)}^{2}+3{x}^{2}{\left(1-x\right)}^{3}\\ -3{\left(1-x\right)}^{2}-3x{\left(1-x\right)}^{2}-3{x}^{2}{\left(1-x\right)}^{2}-3{x}^{3}{\left(1-x\right)}^{2}\\ =\left(1-2x\right)\left[3{x}^{2}{\left(1-x\right)}^{2}-2\right]=\left(1-2x\right){g}_{3}\left( x \right)\end{array}$

$\begin{array}{c}{{f}^{\prime }}_{k+1}\left(x\right)={\left[\left(1+x+{x}^{2}+\cdots +{x}^{k+1}\right)\left(1+\left(1-x\right)+{\left(1-x\right)}^{2}+\cdots +{\left(1-x\right)}^{k+1}\right)\right]}^{\prime }\\ ={\left[{f}_{k}\left(x\right)+\left(1+x+\cdots +{x}^{k}\right){\left(1-x\right)}^{k+1}+\left(1+\left(1-x\right)+\cdots +{\left(1-x\right)}^{k}\right){x}^{k+1}+{x}^{k+1}{\left(1-x\right)}^{k+1}\right]}^{\prime }\\ ={{f}^{\prime }}_{k}\left(x\right)+{H}_{k}\left(x\right)\end{array}$ (8)

$\begin{array}{c}{H}_{k}\left(x\right)=\left(1+2x+\cdots +k{x}^{k-1}\right){\left(1-x\right)}^{k+1}-\left(1+2\left(1-x\right)+\cdots +k{\left(1-x\right)}^{k-1}\right){x}^{k+1}\\ -\left(k+1\right)\left(1+x+\cdots +{x}^{k}\right){\left(1-x\right)}^{k}+\left(k+1\right)\left(1+\left(1-x\right)+\cdots +{\left(1-x\right)}^{k}\right){x}^{k}\\ +\left(k+1\right){x}^{k}{\left(1-x\right)}^{k}\left(1-2x\right)\\ =\left(1-2x\right)\left[{\left(1-x\right)}^{k}+{\left(1-x\right)}^{k-1}x+\cdots +{x}^{k}\right]\\ +2x\left(1-x\right)\left(1-2x\right)\left[{\left(1-x\right)}^{k-1}+{\left(1-x\right)}^{k-2}x+\cdots +{x}^{k-1}\right]\end{array}$

$\begin{array}{c}+\cdots +k{x}^{k-1}{\left(1-x\right)}^{k-1}\left(1-2x\right)\\ -\left(k+1\right)\left(1-2x\right)\left[{\left(1-x\right)}^{k-1}+{\left(1-x\right)}^{k-2}x+\cdots +{x}^{k-1}\right]\\ -\left(k+1\right)x\left(1-x\right)\left(1-2x\right)\left[{\left(1-x\right)}^{k-2}+{\left(1-x\right)}^{k-3}x+\cdots +{x}^{k-2}\right]\\ -\cdots -\left(k+1\right){x}^{k-1}{\left(1-x\right)}^{k-1}\left(1-2x\right)+\left(k+1\right){x}^{k}{\left(1-x\right)}^{k}\left(1-2x\right)\\ =\left(1-2x\right){W}_{k}\left( x \right)\end{array}$

$\begin{array}{l}\left[{\left(1-x\right)}^{k}+{\left(1-x\right)}^{k-1}x+\cdots +{x}^{k}\right]-\left(k+1\right)\left[{\left(1-x\right)}^{k-1}+\cdots +{x}^{k-1}\right]\\ <{x}^{k}-k\left[{\left(1-x\right)}^{k-1}+\cdots +{x}^{k-1}\right]<{x}^{k}-{k}^{2}{x}^{k-1}<-\left({k}^{2}-1\right){x}^{k},\end{array}$

$\begin{array}{l}2x\left(1-x\right)\left[{\left(1-x\right)}^{k-1}+\cdots +{x}^{k-1}\right]-\left(k+1\right)x\left(1-x\right)\left[{\left(1-x\right)}^{k-2}+\cdots +{x}^{k-2}\right]\\ =x\left(1-x\right)\left[2\left({\left(1-x\right)}^{k-2}+{\left(1-x\right)}^{k-3}{x}^{2}+\cdots +{x}^{k-1}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }-\left(k+1\right)\left({\left(1-x\right)}^{k-2}+{\left(1-x\right)}^{k-3}x+\cdots +{x}^{k-2}\right)\right]<0,\end{array}$

$\begin{array}{l}3{x}^{2}{\left(1-x\right)}^{2}\left[{\left(1-x\right)}^{k-2}+\cdots +{x}^{k-2}\right]-\left(k+1\right){x}^{2}{\left(1-x\right)}^{2}\left[{\left(1-x\right)}^{k-3}+\cdots +{x}^{k-3}\right]\\ ={x}^{2}{\left(1-x\right)}^{2}\left[3\left({\left(1-x\right)}^{k-3}+{\left(1-x\right)}^{k-4}{x}^{2}+\cdots +{x}^{k-2}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }-\left(k+1\right)\left({\left(1-x\right)}^{k-3}+{\left(1-x\right)}^{k-4}x+\cdots +{x}^{k-3}\right)\right]<0,\end{array}$

$\cdots ,$

$\begin{array}{l}\left(k-1\right){x}^{k-2}{\left(1-x\right)}^{k-2}\left[{\left(1-x\right)}^{2}+\left(1-x\right)x+{x}^{2}\right]\\ \text{ }-\left(k+1\right){x}^{k-2}{\left(1-x\right)}^{k-2}\left[\left(1-x\right)+x\right]\\ ={x}^{k-2}{\left(1-x\right)}^{k-2}\left[\left(k-1\right)\left(\left(1-x\right)+{x}^{2}\right)-\left(k+1\right)\left(\left(1-x\right)+x\right)\right]<0.\end{array}$

$\begin{array}{c}{W}_{k}\left(x\right)<-\left({k}^{2}-1\right){x}^{k}+\left(k+1\right){x}^{k}{\left(1-x\right)}^{k}\\ <-\left(k+1\right)\left(k-2\right){x}^{k}.\end{array}$ (9)

${{f}^{\prime }}_{k+1}\left(x\right)=\left(1-2x\right)\left[{g}_{k}\left(x\right)+{W}_{k}\left(x\right)\right]=\left(1-2x\right){g}_{k+1}\left(x\right),$

${g}_{k+1}\left(x\right)={g}_{k}\left(x\right)+{W}_{k}\left(x\right)<{W}_{k}\left(x\right)<-\left(k+1\right)\left(k-2\right){x}^{k}<0,\text{}0

${f}_{p}\left(x\right)\ge {f}_{p}\left(\frac{1}{2}\right)={\left(2-\frac{1}{{2}^{p}}\right)}^{2}.$

NOTES

*通讯作者。

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