MVT - Showing $(1+x)^r > 1+rx$

Question

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 $x>0$

???

Answer 1

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.

Answer 2

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 $-1 \leq x \leq 0$.

Answer 3

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 $x>-1$, $x\ne 0$.