MVT - Showing $(1+x)^r > 1+rx$ so as the title states I have the following problem where I have to show:
$(1+x)^r > 1+rx$
When
$r>1$ and if $x>0$ or $-1\le x<0$
So im having alot of trouble with the intution for solving inequalties where one thing implies the other, I don't think you have to apply MVT on this problem since the formula did't.
But could I problem the inequality by doing the following:
$f(x)= (1+x)^r -(1+rx)$
if $f'(x)>0$
and $f(x)>f(0)$
Could I say that for $f'(x)>0$ to be larger than 0, i.e increasing.
$(1+x)^r>(1+rx)$ in order for $f'(x)$ to increase for $x>0$
???
 A: Yes, you can prove $f(x)$ has to be positive for positive inputs by first showing that $f'(x)\geq 0$ or $f'(x)>0$ and then evaluating $f(x)$ at $0$ and seeing that $f(0)>0$. 
It is actually the same as using the MVT in disguise:
$$\frac{f(x)-f(0)}{x-0} = f'(c)$$
for positive numbers $x$ and $c \in (0,\infty)$.
Therefore, if $f'(x)$ is positive for $x>0$, you get:
$$\frac{f(x)-f(0)}{x-0}>0$$
since $x>0$, you conclude that:
$$f(x)-f(0)>0 \implies f(x) > f(0)$$
The same reason works if $f'(x) \geq 0$.
Therefore, your reasoning is correct but you are doing nothing new. You are just covering your application of the MVT under snow. You can also generalize this to a number other than $0$ by noticing that $x \geq a \iff x-a \geq 0$.
Now, take $f(x)=(1+x)^r - (1+rx)$ as you did.
A: Let $f(x)$ be : 
$$f(x) =(1+x)^r - (1+rx)$$
Then : 
$$f'(x) = r(1+x)^{r-1} -r=r[(1+x)^{r-1}-1]$$
This is a continuous function  for $r>1$ and $-1 \leq x \leq 0$, also differentiable, thus we can apply the Mean Value Theorem at $[-1,x]$ :
$$f'(ξ)= \frac{f(x) - f(-1)}{x+1} \Leftrightarrow f'(ξ)(x+1)+f(-1)=f(x) $$
$$\Rightarrow $$
$$f(x) = r[(1+ξ)^{r-1}-1](x+1)+[-(1-r)] $$
$$\Leftrightarrow $$
$$f(x) =  r[(1+ξ)^{r-1}-1](x+1) +r-1 $$
But for $r>1$ and $-1\leq x \leq 0$, it is :
$$ r[(1+ξ)^{r-1}-1](x+1)\geq0$$
$$r-1 > 0$$
Thus $f(x)>0 \Rightarrow (1+x)^r > 1+ rx$ for $r>1$ and $-1 \leq x \leq 0$.
A: You just have to determine the monotonicity of  $f$ on $(-1,+\infty)$.


*

*If $x>0$, $f'(x)=r\bigl((1+x)^{r-1}-1\bigr)>0$ since $r-1>0$ and $1+x>1$, so $(1+x)^{r-1}>1$.

*If $-1<x<0$ $f'(x)<0$ since $0<1+x<1$.


Thus the variations of $f$ are the following:
$$\begin{array}{c|cccc}
x&-1&&0& \\
\hline&||\\
f(x)&||&\searrow&&\nearrow&\\
&||&&0
\end{array} $$
We conclude from these variations that $f(x)>0$ for all $x>-1$, $x\ne 0$.
