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I am often asked to prove inequalities with natural logarithms. An example question I am stuck on is this:

Let $f(x) = \ln(\frac{1+x}{1-x})$ for $-1 < x < 1$. Prove that

$$\frac{x(2+x)}{1+x} \leq f(x) \leq \frac{x(2-x)}{1-x}$$ with equality iff $x=0$.

My general approach has been to try to find a function which upper and lower bounds the function $1/x$ such that the integral of the function will result in the upper and lower bound. I know that the definition of $\ln(x) = \int_{1}^{x} \frac{1}{t}dt$.

Example I want to prove (1)$f_1(x) < \ln(x) <f_2(x)$ . If I can find (2)$F_1(x) < 1/x < F_2(x)$, then by taking the integral I can get the original inequality. Problem is I find it difficult to come up with the (1) relationship. Espically when it gets complicated.

Are there better ways / tricks that can help me solve these proofs more easily? What would be your approach to this problem?

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2 Answers 2

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Let

$$g(x)=\frac{x(2-x)}{1-x}- \ln\frac{1+x}{1-x} $$ $$g’(x)= \frac{x(x^2-x+2)}{(1-x)^2(1+x)},\>\>\> g’’(x) =\frac{2(3x^2+1)}{(1-x)^3(1+x)^2} $$ Note $g(0)=g’(0)=0$ and $g’’(x)>0$, which means that $g(x)$ is convex with the minimum at $g(0)=0$. Thus, $g(x)\ge 0$, or $$\ln\frac{1+x}{1-x} \le \frac{x(2-x)}{1-x}$$ Similarly, let

$$h(x)=\ln\frac{1+x}{1-x} - \frac{x(2+x)}{1+x} $$ $$h’(x)= \frac{x(x^2+x+2)}{(1-x)(1+x)^2},\>\>\>h’’(x) =\frac{2(3x^2+1)}{(1-x)^2(1+x)^3}$$ Note $h(0)=h’(0)=0$ and $h’’(x)>0$, which means that $h(x)$ is convex with the minimum at $h(0)=0$. Thus, $h(x)\ge 0$, or $$ \frac{x(2+x)}{1+x}\le \ln\frac{1+x}{1-x} $$

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I don't know if you like this method which doesn't need derivatives. Suppose $x\ge 0$. For $t\in[0,x]$,$1-t\ge 1-x,1+t\ge 1$ and hence $$ \ln(\frac{1+x}{1-x})=\int_0^x\bigg(\frac{1}{1+t}+\frac{1}{1-t}\bigg)\le\int_0^x\bigg(1+\frac{1}{1-x}\bigg)dt=\frac{x(2-x)}{1-x}. \tag1$$ Similarly for $t\in[0,x]$,$1-t\ge 1$ $$ \ln(\frac{1+x}{1-x})=\int_0^x\bigg(\frac{1}{1+t}+\frac{1}{1-t}\bigg)\ge\int_0^x\bigg(\frac{1}{1+x}+1\bigg)dt=\frac{x(2+x)}{1+x}. \tag2$$ Combining (1) and (2) gives the desirable inequality. If $x<0$, you can do the same.

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  • $\begingroup$ shoukld the integral you are comparing be with a $t$ and not $x$? $\endgroup$ Nov 25, 2020 at 15:46
  • $\begingroup$ Comparing $t$ and $x$. Let me put more details. $\endgroup$
    – xpaul
    Nov 25, 2020 at 15:48

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