高数强化NO25|不等式证明|函数不等式|常数不等式|含有积分的不等式

函数不等式

$$\begin{array}{rl}& \text{分析：}\\ & \text{化简：}F(x)=(1+x)\ln (1+x)-x\ln x,x>1\\ & \text{要证}F(x)>0.\\ & \text{只需证}F(x)>F(1)=2\ln 2,\\ & \text{只需证}F(x)\nearrow ,\\ & \text{只需证}{F}^{\prime }(x)>0.\\ & F^{\prime }(x)=\ln (1+x)+1-\ln x-1=\ln (1+x)-\ln x>0\end{array}$$
$$\begin{array}{rl}& \text{证明：设}F(x)=(1+x)\ln (1+x)-x\ln x,x\ge 1\\ & F^{\prime }(x)=\ln (1+x)-\ln x>0(x\ge 1)\\ & \text{故}F(x)\text{在}[1,+\infty )\text{内单调递增}.\\ & (1+x)\ln (1+x)-x\ln x>F(1)=2\ln 2>0(x\ge 1)\\ & \\ & \text{整理可得：}\frac{\ln (1+x)}{\ln x}>\frac{x}{1+x}(x>1)\end{array}$$
$$\begin{array}{rl}& \text{【小结】1.不等式证明的基本思路如下：假设要证明}f(x)>g(x)，x\in (a,b)，\text{首先令}\\ & F(x)=f(x)-g(x)，\text{然后计算}F(a)\text{及}F(b)；\text{一般情况下，这两个数中会有一个等于}\\ & \text{零。不妨设}F(a)=0，\text{此时只需证明}F(x)\text{在}[a,b]\text{上单调递增，即只需证明}{F}^{\prime }(x)>0，\\ & x\in (a,b)\text{即可。一般来说，}{F}^{\prime }(x)\text{要比}F(x)\text{简单，如果}{F}^{\prime }(x)>0\text{仍不容易证明，则重}\\ & \text{复前面的过程，一般来说，考研中此类题目的求导次数不会超过两次。}\\ & 2.\text{构造辅助函数之前，要注意先对不等式进行化简，再进行计算。}\end{array}$$

$$\begin{array}{rl}& \text{分析：化简的有两个：①定义域（值域）}\text{②函数表达式}\\ & F(x)=x[\ln (1+x)-\ln (1-x)]+\cos x-1-\frac{x^2}{2},0\le x<1\\ & F(x)\ge 0=F(0)\\ & F(x)\nearrow ,F^{\prime }(x)>0.\\ & F^{\prime }(x)=\ln \frac{1+x}{1-x}+\frac{x}{1+x}+\frac{x}{1-x}-\sin x-x\\ & =\ln \frac{1+x}{1-x}+\frac{2x}{1-x^2}-\sin x-x\\ & >0\end{array}$$
$$\begin{array}{rl}& \text{证明:}\\ & \text{法1: 考场上不会硬凑法}\\ & \text{令}F(x)=x[\ln (1+x)-\ln (1-x)]+\cos x-1-\frac{x^2}{2},-1<x<1\\ & \text{显然, }F(x)\text{为偶函数.}\\ & \text{则在}[0,1)\text{上对}F(x)\text{求导:}\\ & F^{\prime }(x)\\ & =\ln \frac{1+x}{1-x}+\frac{2x}{1-x^2}-\sin x-x\\ & >\ln \frac{1+x}{1-x}+\frac{2x}{1-x^2}-x-x\\ & =\ln \frac{1+x}{1-x}+\frac{2x^3}{1-x^2}\\ & >0\\ & \text{则}F(x)\text{在}[0,1)\text{上单调递增. }F(x)\ge F(0)=0,x\in [0,1)\\ & \text{又}F(x)\text{为偶函数, 故}F(x)\text{在}(-1,1)\text{上满足}F(x)\ge 0\\ & \text{即}x\in (-1,1)\text{时, }x\ln \frac{1+x}{1-x}+\cos x\ge 1+\frac{x^2}{2}.\end{array}$$
$$\begin{array}{rl}\text{法2：}{F}^{\prime \prime }(x)& =\frac{1}{1+x}+\frac{1}{1-x}+\frac{2}{1+x^2}+2x\cdot \frac{2x}{(1+x^2{)}^2}-\cos x-1\\ & \ge 0+0+2+0-1-1=0\end{array}$$
$$\begin{array}{rl}& \text{【小结】若辅助函数在两个端点的函数值均不为零，则一般需要在区间内部找一个等于}\\ & \text{零的点，然后再将区间拆成两段之后再分别证明。}\end{array}$$

$$\begin{array}{rl}& \text{分析:}\\ & F(x)=(x-1)\left(x-{\ln }^{2}x+2k\ln x-1\right)\ge 0\\ & \\ & \text{当}0<x<1\text{时,}\\ & G(x)=x-{\ln }^{2}x+2k\ln x-1\le 0=G(1)\\ & G(x)\nearrow ,G^{\prime }(x)>0\\ & \\ & \text{当}x>1\text{时,}\\ & G(x)=x-{\ln }^{2}x+2k\ln x-1\ge 0=G(1)\\ & G(x)\nearrow ,G^{\prime }(x)>0\end{array}$$
$$\begin{array}{rl}& h(x)=x-2\ln x+2k>0\\ & h^{\prime }(x)=1-\frac{2}{x}=0,x=2\\ & (0,2):h(x)\searrow x>2:h\nearrow h(x)\text{在}x=2\text{取最小值.}\end{array}$$
$$\begin{array}{rl}& \text{证：令}G(x)=x-{\ln }^{2}x+2k\ln x-1,x>0\\ & G^{\prime }(x)=1-2\ln x\cdot \frac{1}{x}+\frac{2k}{x}\\ & =\frac{x-2\ln x+2k}{x}\\ & \\ & \text{令}h(x)=x-2\ln x+2k,x>0\\ & h^{\prime }(x)=1-\frac{2}{x},\text{则}x\in (0,2),h^{\prime }(x)<0,h(x)\text{单调递减；}\\ & x\in (2,+\infty ),h^{\prime }(x)>0,h(x)\text{单调递增.}\\ & h(x)\text{在}x=2\text{处取得最小值.}\\ & h(x)\ge h(2)=2-2\ln 2+2k\ge 0\\ & \\ & \text{故}{G}^{\prime }(x)\ge 0,G(x)\nearrow ,G(1)=0.\\ & \text{则}(x-1)G(x)\ge 0,\text{命题得证.}\end{array}$$
$$\begin{array}{rl}& \text{【小结】如果辅助函数在两个端点处的函数值均不为零，且区间内部的零点不容易找到，}\\ & \text{则可以先求导，根据函数的单调性找到函数的最小值点，再验证最小值点处的函数值非}\\ & \text{负即可。}\end{array}$$

$$记住结论：0<t\le \frac{\pi }{2}，t>\sin t>\frac{2t}{\pi }$$
$$\begin{array}{rl}& \text{分析:}\\ & F(x)=\sin \frac{x}{2}-\frac{x}{\pi },x\in [0,\pi ]\\ & F(x)>0,F(x)>F(0),\text{当两个端点均为0时}\\ & \text{不能再用单调性,}\\ & \text{要用凹凸性.}\end{array}$$
$$\begin{array}{rl}& \text{证：令}F(x)=\sin \frac{x}{2}-\frac{x}{\pi },0\le x\le \pi \\ & F^{\prime }(x)=\frac{1}{2}\cos \frac{x}{2}-\frac{1}{\pi }\\ & F^{\prime \prime }(x)=-\frac{1}{4}\sin \frac{x}{2}<0,0\le x\le \pi \\ & \text{故}F(x)\text{在}[0,\pi ]\text{上为凸函数.}\\ & \text{又}F(0)=F(\pi )=0,\text{则}\\ & F(x)>0.\text{整理可得}\sin \frac{x}{2}>\frac{x}{\pi }\end{array}$$
$$\begin{array}{rl}& \text{【小结】如果辅助函数在两个端点的值均为零，则可以考虑凹凸性：如果为凸函数，则}\\ & \text{函数大于零；如果为凹函数，则函数小于零。}\end{array}$$

常数不等式

$$\begin{array}{rl}& \text{分析:}\\ & \text{右边:}\\ & \ln b-\ln a<\frac{b-a}{\sqrt{ab}}\\ & \\ & ①\text{令}F(x)=\ln x-\ln a-\frac{x-a}{\sqrt{ax}}\\ & F(b)<0,F(b)<F(a),F(x)\searrow ⟹F^{\prime }(x)<0\\ & \\ & ②\text{令}F(x)=\ln b-\ln x-\frac{b-x}{\sqrt{bx}}\\ & F(a)<0,F(a)<F(b),F(x)\nearrow ⟹F^{\prime }(x)>0\\ & \\ & ③\text{令}F(b)=\ln b-\ln a-\frac{b-a}{\sqrt{ab}}\\ & F(b)<0,F(b)<F(a),F(b)\searrow ⟹F^{\prime }(b)<0\\ & F^{\prime }(b)=\frac{d(F(b))}{d(b)}\end{array}$$
$$\text{左边:}F(x)=2a(x-a)-(a^2+x^2)(\ln x-\ln a),x\ge a$$
$$\begin{array}{rl}& \frac{2ab-a^2}{a^2+b^2}<\ln b-\ln a=\ln \frac{b}{a}\\ & \frac{\frac{2b}{a}-1}{1+{\left(\frac{b}{a}\right)}^2}<\ln \frac{b}{a},2\cdot \frac{b}{a}-1<\ln \frac{b}{a}\cdot \left(1+{\left(\frac{b}{a}\right)}^2\right)\end{array}$$
$$\text{令}F(x)=2x-1-(1+x^2)\ln x$$
$$\begin{array}{rl}& \text{证：先证左边.}\\ & \text{令}F(x)=2x-1-(1+x^2)\ln x,x>1\\ & F^{\prime }(x)=2-2x\ln x-\frac{1+x^2}{x}\\ & =\frac{2x-x^2-2x^2\ln x-1}{x}\\ & =\frac{-(x-1{)}^2-2x^2\ln x}{x}<0\\ & \text{故}F(x)\searrow ,F(x)<F(1)=0\\ & F\left(\frac{b}{a}\right)<0,\text{即}2\cdot \frac{b}{a}-1-\left(1+{\left(\frac{b}{a}\right)}^2\right)\ln \frac{b}{a}<0\\ & \text{整理可得}\frac{2a}{a^2+b^2}<\frac{\ln b-\ln a}{b-a}\end{array}$$
$$\begin{array}{rl}& \text{证：}\\ & \text{令}F(x)=2a(x-a)-(a^2+x^2)(\ln x-\ln a),x>a\\ & F^{\prime }(x)=2a-2x(\ln x-\ln a)-\frac{a^2+x^2}{x}\\ & =\frac{2ax-2x^2(\ln x-\ln a)-a^2-x^2}{x}\\ & =\frac{2x(\ln a-\ln x)-(a-x{)}^2}{x}\\ & <0\end{array}$$
$$\begin{array}{rl}& \text{证：令}F(x)=\ln x-\ln a-\frac{x-a}{\sqrt{ax}},x>a.\\ & F^{\prime }(x)=\frac{1}{x}-\frac{1}{2\sqrt{a}\cdot \sqrt{x}}+\sqrt{a}\cdot \left(-\frac{1}{2}\right)x^{-\frac{3}{2}}\\ & =\frac{1}{x}-\frac{1}{2\sqrt{a}\cdot \sqrt{x}}-\frac{\sqrt{a}}{2x\sqrt{x}}\\ & =\frac{2\sqrt{a}\cdot \sqrt{x}-x-a}{2\sqrt{a}\cdot x\sqrt{x}}=\frac{-(\sqrt{x}-\sqrt{a}{)}^2}{2ax\sqrt{x}}<0\\ & \text{故}F(x)\searrow ,F(a)=0,\text{因此}F(x)<F(a)=0,x>a.\\ & \text{整理可得}\frac{\ln b-\ln a}{b-a}<\frac{1}{\sqrt{ab}}\end{array}$$
$$\begin{array}{rl}& \text{【小结】1.证明常数不等式基本思路是常量变量化，即把不等式的一部分常数写成}x，\\ & \text{将其看成函数不等式，再利用前面的方法证明。}\\ & 2.\text{不等式中如果出现了两个}f\text{相减的结构，也可以考虑使用拉格朗日中值定理进行变形。}\end{array}$$

$$\begin{array}{rl}& \text{证：}y=f(x)\text{在}(b,f(b))\text{处的切线方程为}\\ & Y-f(b)=f^{\prime }(b)(X-b),\text{令}Y=0,\text{则}{X}_0=b-\frac{f(b)}{f^{\prime }(b)}.\\ & f^{\prime }(x)>0,\text{故}f(b)>f(a)=0.\text{又}{f}^{\prime }(b)>0,\text{故}{X}_0<b.\\ & \\ & \text{在}(a,b)\text{内用拉格朗日中值定理：}\frac{f(b)-f(a)}{b-a}=f^{\prime }(\xi ),\xi \in (a,b).\\ & \text{因为}{f}^{\prime \prime }(x)>0,\text{故}{f}^{\prime }(x)\nearrow ,f^{\prime }(\xi )<f^{\prime }(b).\\ & \\ & \text{整理可得}b-\frac{f(b)}{f^{\prime }(b)}>a，综上，a<x_0<b\end{array}$$

含有积分的不等式

$$\begin{array}{rl}& \text{解: (1) 因为}0\le g(x)\le 1,0\le {\int }_a^{x}g(t)dt\le {\int }_a^{x}1dt=x-a\\ & \\ & (2)\text{令}F(x)={\int }_a^{a+{\int }_a^{x}g(t)dt}f(u)du-{\int }_a^{x}f(t)g(t)dt\\ & F^{\prime }(x)=f\left(a+{\int }_a^{x}g(t)dt\right)\cdot g(x)-f(x)g(x)\\ & =g(x)\left(f\left(a+{\int }_a^{x}g(t)dt\right)-f(x)\right)<0\\ & \left(g(x)\ge 0,a+{\int }_a^{x}g(t)dt\le x,f(x)\nearrow \right)\\ & F(x)\text{单调递减, }F(b)\le F(a),\text{整理可得}{\int }_a^{a+{\int }_a^{b}g(t)dt}f(x)dx\le {\int }_a^{b}f(x)g(x)dx\end{array}$$
$$\begin{array}{rl}& \text{【小结】对积分不等式，证明的基本思路也是把它看成常数不等式，这里最常用的证明}\\ & \text{方法一般是常量变量化，将其化作函数不等式来证明。}\end{array}$$

$$\begin{array}{rl}& \text{证：}\\ & {\int }_a^{b}xf(x)dx={\int }_a^{b}x\cdot {\left({\int }_a^{x}f(t)dt\right)}^{\prime }dx={x{\int }_a^{x}f(t)dt|}_a^b-{\int }_a^b\left({\int }_a^{x}f(t)dt\right)dx\\ & =b{\int }_a^{b}f(t)dt-{\int }_a^b\left({\int }_a^{x}f(t)dt\right)dx\\ & \\ & {\int }_a^{b}xg(x)dx={\int }_a^{b}x\cdot {\left({\int }_a^{x}g(t)dt\right)}^{\prime }dx={x{\int }_a^{x}g(t)dt|}_a^b-{\int }_a^b\left({\int }_a^{x}g(t)dt\right)dx\\ & =b{\int }_a^{b}g(t)dt-{\int }_a^b\left({\int }_a^{x}g(t)dt\right)dx\\ & \\ & \text{故}{\int }_a^{b}xf(x)dx={\int }_a^{b}xg(x)dx\end{array}$$