# Prove an inequality (Using Taylor expansion)

Prove: $\frac{x}{1+x} < \ln(1+x) < x$.

I thought a good practice would be to prove it using Taylor Expansion.

Here's my try:
$$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3}...$$

The n=1 Taylor polynomial is: $$T_1(x) = x$$ and $$ln(1+x) = T_1(x) + R_1(x)$$

Lets evaluate $R_1(x)$ by Cauchy's remainder formula:

$$R_1(x) = \frac{f^{(2)}(\xi)}{2!}\cdot x^2 = \frac{\frac{-1}{(\xi+1)^2}}{2!}\cdot x^2 = \frac{-x^2}{2(\xi+1)^2} < 0$$

Now, it does prove the right-hand side because $x + R_1(x) < x$ ($R_1(x)$ is negative). I'm not so sure what should I do for the left-hand side. I'd also like to get general critique for my current work.

Thanks!

We apply the mean value theorem on the function $t\mapsto \ln t$ on the interval $[1,1+x]$: there's $\zeta\in(1,1+x)$ such that $$\ln(1+x)=\frac x\zeta$$ and notice that $$\frac1{1+x}<\frac1\zeta<1$$

• Very nice, thank you! – AnnieOK Jun 22 '14 at 16:00
• You're welcome. – user63181 Jun 22 '14 at 16:02

Using the elementary inequality $$1+x\le e^x$$ one directly obtains one side of the inequality chain. Replace $x$ by $-y$ and invert to obtain $$\frac1{1-y}\ge e^y$$ and then set $1+x=\frac1{1-y}=1+\frac{y}{1-y}$ or $y=\frac{x}{1+x}$ to obtain $$1+x\ge e^{\frac{x}{1+x}}$$ for the other part of the inequality chain.

Proof:

Assuming you argued and justified that $$f (x) = Ln (1 + x)$$ is continuous and derivable $$\forall x> 0$$.

This is equivalent to writing the interval $$(0, x)$$.

By the Mean Value theorem $$\exists c \in (0,x)$$ such that $$f'(c)=\frac{f(b)-f(a)}{b-a}$$, i.e. $$\frac{1}{1+c}=\frac{Ln(1+x)-Ln(1)}{x-0}=\frac{Ln(1+x)}{x}$$.

We can delimit $$\frac{1}{1+c}=f'(c)$$ (*), then:

$$\frac{1}{1+x}

But we know that $$f'(c)=\frac{1}{1+c}=\frac{Ln(1+x)}{x}$$ (**)

Replacing in (*) we get:

$$\frac{1}{1+x}<\frac{Ln(1+x)}{x}<1$$

Multiplying the inequality by $$x$$, we obtain the desired result:

$$\frac{x}{1+x}.