# Prove Inequalities $\frac{x(2+x)}{1+x} \leq f(x) \leq \frac{x(2-x)}{1-x}$ Using Calculus

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)$$f_1(x) < \ln(x) . If I can find (2)$$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?

$$g(x)=\frac{x(2-x)}{1-x}- \ln\frac{1+x}{1-x}$$ $$g’(x)= \frac{x(x^2-x+2)}{(1-x)^2(1+x)},\>\>\> g’’(x) =\frac{2(3x^2+1)}{(1-x)^3(1+x)^2}$$ 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 $$\frac{x(2+x)}{1+x}\le \ln\frac{1+x}{1-x}$$
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.
• shoukld the integral you are comparing be with a $t$ and not $x$? Nov 25, 2020 at 15:46
• Comparing $t$ and $x$. Let me put more details. Nov 25, 2020 at 15:48