# Proving an inequality which the solution says “is obvious” but I cant see how

I was going through my Lebesgue Measure and Integration course, and I came across this inequality. $$\left(1+\frac{x}{n}\right)^{-n}\leq e^{-x}$$
I tried expanding both sides, and got
$$\text{LHS}= 1-x+\frac{n+1}{n}\frac{x^2}{2!}-\frac{(n+1)(n+2)}{n^2}\frac{x^3}{3!}\cdots$$
and $$\text{RHS}=1-x+\frac{x^2}{2!}\cdots$$ Now, I noticed that all the terms in the LHS are either equal or bigger than the corresponding terms in RHS but I don't know how to conclude the inequality from this. Am I missing something extremely simple and obvious here?
Any help will be appreciated.

• I'd recommend taking the derivative of $(1+x/n)^n - e^x$ after having taken inverses on both sides (be careful if you do take inverses of both sides) – Max Feb 6 '17 at 15:29
• The inequality is in reverse and holds for non-negative values of $x$ and positive integer values of $n$. – Paramanand Singh Feb 6 '17 at 15:35
• @ParamanandSingh Forgot to put in that $x$ is non-negative. – S.C.B. Feb 6 '17 at 15:43
• Sometimes an author might say something is "obvious" merely because it follows immediately from something that the author assumes the reader knows. – Michael Hardy Feb 6 '17 at 16:04

For your inequality to be true, we need $n<0$ and $x \ge 0$. Note that from this link we know that $$e^{\frac{x}{n}} \ge \;1+\frac{x}{n}$$ Now, as $t^{-n}$ is an increasing function if $n<0$, $t>0$, we exponentiate each side by $-n$ to get $$e^{-x} \ge \bigg{(}1+\frac{x}{n}\bigg{)}^{-n}$$ as desired. Of course, if $n \ge 0$, the inequality signs reverse, for a similar reason. So your inequality is, for example, not true when $n=1$ as stated previously.

• Let me see if there's any restriction.. – Sum-Meister Feb 6 '17 at 15:28
• @Sum-Meister Okay. I will edit accordingly if so. – S.C.B. Feb 6 '17 at 15:29
• Rearranging the original inequality, we get: $e^x \leq (1+\frac{x}{n})^n$, which is, as we all know, false for all n, since this series is increasing and converges to $e^x$. – Omry Feb 6 '17 at 15:31
• The other direction is true though. – Omry Feb 6 '17 at 15:32
• @Sum-Meister How long is this going to take? Not trying to hurry you on, but it does seem to be taking some time. Just going to ask when. – S.C.B. Feb 6 '17 at 15:51

We can take a logarithm from both sides. Then, we have $-n \log(1 + \frac{x}{n}) \leq -x \Rightarrow n \log(1 + \frac{x}{n}) \geq x \Rightarrow \log(1 + \frac{x}{n}) \geq \frac{x}{n}$

let $\frac x n = y$. Then, we have

$\log(1+y) \geq y$ and this is not true for any $y$