<s>[INST] 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)$f_1(x) < \ln(x) <f_2(x)$ . 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?
 [/INST] 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
$$ \frac{x(2+x)}{1+x}\le \ln\frac{1+x}{1-x}  $$
</s>