Prove that $$x-\frac{x^2}{2}\le ln(x+1)\le x$$ for $$x\ge 0$$

Here is what I've done so far : $1-\frac{x}{2} \le \frac{ln(1+x)}{x} \le 1$ then from MVT we have $f(x)=ln(x)$ in that case we have $\frac{ln(1+x)}{x}=f'(c)=\frac{1}{c}$ And that said $\frac{1}{c}\le 1$ ($c\in(1,1+x)$) But i just can't prove the other half which is $1-\frac{x}{2}\le \frac{ln(x+1)}{x}$

It's due for today so any help would be appreciated, Thanks in advance.

  • $\begingroup$ I did not quite follow your reasoning for part 1, though I would agree that if $f(x) = \log(1+x)$ then the MVT says for $x>0$ that $\frac{f(x)-f(0)}{x} = f'(c)$ for some $c \in [0, x]$. For part 2, what if you define $g(x) = x-x^2/2$ and compare derivatives of $g$ and $\log(1+x)$? $\endgroup$ – Michael Oct 31 '16 at 20:05

Take $g(x) = \ln(x+1) - x + \frac{x^2}{2}$. Then from the Mean Value Theorem we have that $\exists c \in (0,x)$, s.t.

$$g'(c) = \frac{g(x) - g(0)}{x-0} \implies \frac1{c+1} - 1 + c = \frac{\ln(x+1) - x + \frac{x^2}{2}}{x}$$

Now it's enough to prove that the LHS is positive, which is true as:

$$\frac1{c+1} - 1 + c > 0 \iff \frac{c^2+c-c-1+1}{c+1} > 0 \iff \frac{c^2}{c+1} > 0$$

But this is obviously true, as $c> 0$


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