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I am currently trying to prove the following relationship $$1 + r \leq \left(1 +\frac{r}{m}\right)^m\quad \text{for any }m \geq 1.$$

Would you be so kind and provide some hints/solutions to the above?

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    $\begingroup$ You might study the function $f(x) = e^{x \ln \left(1+\frac{r}{x}\right)}$. $\endgroup$ – Joel Cohen Apr 23 '12 at 20:56
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    $\begingroup$ This is simply Bernoulli Inequality $\endgroup$ – TenaliRaman Apr 23 '12 at 21:19
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You'll probably want to assume at least $r \ge -m$, otherwise it can be false (e.g. try $m=3$ and $r < -9$).

Let $f(r) = (1+r/m)^m - (1+r)$ for $-m \le r < \infty$. Then $f(0) = f'(0) = 0$, while $f''(r) = \frac{m-1}{m} (1+r/m)^{m-2} \ge 0$, so $f(r)$ is convex on this interval. Therefore $f(r) \ge 0$ there.

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  • $\begingroup$ It might be the hangover, but $m=4,r=-12$ gives $1-12=-11\leq\left(1-\dfrac{12}{4}\right)^4=16$ which is true. $\endgroup$ – Asaf Karagila Apr 24 '12 at 6:25
  • $\begingroup$ @AsafKaragila: Of course the inequality is true when $r \le -1$ and $m$ is an even integer (left side $\le 0 \le$ right side). It is false if $m$ is an odd integer $\ge 3$ and $r < -3m$, and it is meaningless if $r < -m$ for most non-integers $m$. $\endgroup$ – Robert Israel Apr 24 '12 at 6:59
  • $\begingroup$ So it is the hangover! :-) $\endgroup$ – Asaf Karagila Apr 24 '12 at 7:12
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This is not a very nice solution, but for $m$ not an integer:

It is easy to see that $\frac{\log(1+x)}{x}$ times a constant is a decreasing function (simply because $(\log(1+x)≤ x)$ and x grows faster). Hence for $x=\frac{r}{m}$ the function $$ m*\log(1+r/m) $$ is increasing for increasing $m≥1$. But therefore so is the function $(1+\frac{r}{m})^m$.

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  1. If you do not know $\left(a+b\right)^n=\sum_{i=0}^n\binom{n}{i}a^ib^{n-i}$, prove it by induction using $\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$.

  2. Note that $\binom{m}{0}=1$, $\binom{m}{1}=m$ for every $m$, and use it to prove the inequality.

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