so as the title states I have the following problem where I have to show:

$(1+x)^r > 1+rx$


$r>1$ 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 $x>0$



3 Answers 3


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:


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.

  • $\begingroup$ So what would be sufficent answer? Because $f(x)=(1+x)^r - (1+rx)$ is not equal to zero I can't prove the inequality just by moving $- (1+rx)$. Which gives $(1+x)^r > (1+rx)$. Could I say that $f(x)=(1+x)^r - (1+rx)$ implies $(1+x)^r > (1+rx)$? Thank you btw $\endgroup$ Dec 1, 2017 at 1:35
  • $\begingroup$ @AliasaZarownyPseudonymia: I'm not sure I have understood your question. You can add and subtract terms from both sides of an inequality. But you CANNOT multiply or divide terms unless you know for sure that they are either positive or negative. In case of multiplying or dividing by a negative term, the direction of the inequality is reversed. But there is no problem with adding or subtracting things from both sides of an inequality. Is that what you were asking? $\endgroup$ Dec 1, 2017 at 1:41
  • $\begingroup$ I was alluding to the inequality $f(x)>f(0)$. I was wondering how to end the proof. If you substitute $f(x)$ with it's original equation and the same thing for $f(0)$. You'd get $(1+x)^r -(1+rx)>f(0)=1^r-1$. Since the RHS for $f(x)>f(0)$ does't equal zero I could not simply just add $(1+rx)$ to both sides in order to get the inequality $(1+x)^r>(1+rx)$, which is what ii'm trying to prove. So how would I phrase my final answer? $\endgroup$ Dec 1, 2017 at 1:58
  • $\begingroup$ @AliasaZarownyPseudonymia: Well, isn't $1^r=1?$ for any exponent $r$? Therefore, $f(0)=0$? But, please note that proving it for $-1 \leq x < 0$ is a bit trickier. I didn't cover it in my answer word by word but it's essentially the same idea. $\endgroup$ Dec 1, 2017 at 2:02
  • $\begingroup$ You're right, major oversight on my behalf. Thank you! $\endgroup$ Dec 1, 2017 at 2:05

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 $-1 \leq x \leq 0$.

  • $\begingroup$ Should it however be $f(x)\ge 0 $ instead for $f(x)> 0 $ considering that $$ r[(1+ξ)^{r-1}-1](x+1)\geq0$$? $\endgroup$ Dec 1, 2017 at 0:54

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


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