Showing $(1+x)^r<(1+rx)$ (MVT) [duplicate]

So the full question goes as following:

Let $0<r<1$.

If $x>0$ or $-1\le x<0$, show that $(1+x)^r < (1+rx)$

I'm having a real hard time getting the idea of the usefullness of MVT. It seems every approach I take is faulty.

I don't know how to think. If I take $f'(c)$ for $0<c<x$ I get $(1+c)r -1 / c$ which I can conclude is larger than $0$, which leads me nowhere.

Help is very appriciated!

just a hint

Let $F (x)=(1+x)^r-1-rx$

then $$F (x)=F (0)+xF'(c)$$

with

$$F'(c)=r ((1+c)^{r-1}-1)$$ $$=rc (r-1)(1+c')^{r-2}$$

Observe that $x$ and $c$ have the same sign.

• Alright so I got: $(1+x)^r = rx((1+c)^{r-1}+(1+rx)$ So im trying to prove $rx((1+c)^{r-1}$ is smaller than 1 essentially? – Aliasa Zarowny Pseudonymia Dec 1 '17 at 19:29
• @AliasaZarownyPseudonymia You can use MVT a second time. – hamam_Abdallah Dec 1 '17 at 19:35
• I noticed however for $(1+x)^r = rx((1+c)^{r-1}-1)+(1+rx)$ that $rx((1+c)^{r-1}-1)$ will always be negative since $(1+c)^{r-1}-1) = \frac{1}{(1+c)^{r-1}}-1 <0$ because $rx>0$ $rx((1+c)^{r-1}-1)<0$. Is that a legitimate reasoning? Or am I doing something wrong? – Aliasa Zarowny Pseudonymia Dec 1 '17 at 19:44

By any of the answers to your previous question (sma question, case $r>1$), we can set $r=\dfrac 1s,\; (s>1)$, we have \begin{align} &&(1+rx)^s&=\Bigl(1+\frac xs\Bigr)^s>1+s\cdot\frac xs=1+x\\ &\text{whence}&1+rx&>(1+x)^{\tfrac1s}=(1+x)^r. \end{align}

Let $f(x)=(1+x)^r$ and then $f'(x)=r(1+x)^{r-1}$. Then by the MVT, $$f(x)-f(0)=f'(c)x, \text{ for some c between 0 and x}. \tag{1}$$ If $x>0$, then $$f'(x)=r(1+x)^{r-1}<r$$ and hence from (1), one has $$(1+x)^r-1=f'(c)x<rx$$ or $$(1+x)^r<1+rx.$$ If $x\in(-1,0)$, then $0<1+x<1$ and hence $f'(x)=r(1+x)^{r-1}>r$. So $$(1+x)^r-1=f'(c)x<rx$$ or $$(1+x)^r<1+rx.$$ for $x\in(-1,0)$. Thus if $x\in(-1,0)\cup(0,\infty)$, one has $$(1+x)^r<1+rx.$$

• thanks it was very easy to follow. However why the derivative of f'(x) pertinent, I see that you used it to find a relationship between $r(1+x)^{r-1}$ and $r$. But what does it mean? I know that usually it is to confirm whether the function is increasing from the starting point $f(0)$ or not. If it is increasing you can conclude that $f(x)>f(0)$ but what does $r(1+x)^{r-1}<r$ tell me? Thank you btw – Aliasa Zarowny Pseudonymia Dec 1 '17 at 19:59
• @AliasaZarownyPseudonymia, if $x>0$, then $1+x>1$ and hence $(1+x)^{r-1}<1$ since $r<1$, – xpaul Dec 1 '17 at 20:24
• I forumlated myself poorly; what is the rational of taking the derivative $f'(x)$? – Aliasa Zarowny Pseudonymia Dec 1 '17 at 20:35
• How does $f'(x)<r$ imply $f'(c)<r$? (Sorry for bombarding you with questions) – Aliasa Zarowny Pseudonymia Dec 1 '17 at 20:52
• @AliasaZarownyPseudonymia, $(1+x)^{r-1}<1$ implies $f'(x)=r(1+x)^{r-1}<r$ for $x>0$ and hence $f'(c)<r$. – xpaul Dec 1 '17 at 21:25