Prove that $1+rx = (1+x)^r$ holds only if $x=0$.

Question

I believe my work is correct here, just looking for closure. Is it true  
that since r>1, the last line could be (1+x)^1?

Suppose $1+rx\le(1+x)^r$ for any real number $r>1$ and $x>-1$. Prove that the equality holds only if $x=0$.

My work: For $x=0$, letting $f(x) = (1+x)^r$: [mean value theorem]

$f'(c) = \frac{f(x)-f(0)}{x-0}$

$r(1+c)^{r-1} = \frac{(1+x)^r-1^r}{x-0}$

$\frac{(1+x)^r-1^r}{(1+c)^{r-1}}=xr$

$(1+x)-1=xr$

$1+x = 1+xr$

I don't follow how your 3rd line implies your 4th. – Christopher A. Wong Mar 18 '13 at 18:18 — Christopher A. Wong, Mar 18 '13 at 18:18
Since r>1, r-1 should cancel r to at least 1 right? – user65384 Mar 18 '13 at 18:24 — user65384, Mar 18 '13 at 18:24

score 2 · Answer 1 · answered Mar 18 '13 at 18:20

2

Let $f(x)=(1+x)^r-rx-1$ then $f'(x)=r(1+x)^{r-1}-r>0$ for $x\neq 0$ since $r-1>0$, so $f$ is monotonically increasing and $f(0)=0$ hence $f(x)\neq 0$ for all $x\neq 0$. We can conclude.

answered Mar 18 '13 at 18:20

Prove that $1+rx = (1+x)^r$ holds only if $x=0$.

1 Answers1