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How do you prove $e^{x} \geq \left(1+\frac{x}{n}\right)^{n}$ for $n \geq 1$? I can prove this for natural numbers only via induction, but how do you prove this for any real $n \geq 1$?

We start with the base case $n=1$. We have $e^x \geq 1+x$ by a variety of methods. For the induction step, assume $e^{x} \geq \left(1+\frac{x}{n}\right)^{n}$. Notice that taking the derivative of $(1+\frac{x}{n+1})^{n+1}$ gives us $(1+\frac{x}{n+1})^{n}$ and thus $(1+\frac{x}{n+1})^{n} < \left(1+\frac{x}{n}\right)^{n} \leq e^x = \frac{d}{dx} e^x$.

I'm not sure how to extend this to the non-integer case. Any help would be appreciated.

Snowball
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    What is your definition of $x^n$ for real $n \ge 1$? For irrational $n$, there is no "intuitive" definition any more; you have to be precise about what your definition is in order to prove this. – Misha Lavrov Jun 29 '21 at 21:14
  • I'm not sure I buy the inductive argument; it is not true that $f'(x) \leq g'(x) \Rightarrow f(x) \leq g(x)$ although maybe this is not what you are saying. – hunter Jun 29 '21 at 21:18
  • This is equivalent to $x\geq n\left(1+\frac xn\right),$ so you just need t9 know that $\log (1+y)\geq y$ for any $y.$ – Thomas Andrews Jun 29 '21 at 21:20
  • When $x<-n,$ the right side is not in general defined, so you have to assume $y>-n.$ – Thomas Andrews Jun 29 '21 at 21:21
  • @MishaLavrov You got me there. I vaguely recall in Spivak's Calculus there being a discussion about irrational exponents. But in this context I really just am looking for this in the context of a inequalities cheat sheet I saw to be used in computer science. – Snowball Jun 29 '21 at 21:21
  • Proving $e^x \geq 1 + x$ is enough as you can write the above expression as $e^{\frac{x}{n}} \geq 1 + \frac{x}{n}$. Just set $y = \frac{x}{n}$ and you are done. – Snowball Jun 29 '21 at 21:47

2 Answers2

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In this answer I define the natural logarithm as $$ \log x =\int_{1}^{x}\frac{1}{t} \, dt \, . $$ and the exponential function as the inverse of the logarithm. This leads to the definition $a^x=\exp(x\log a)$ for $a>0$, valid for all real $x$.

We begin with the fact that $$ \frac{x}{n} \geq \log\left(1+\frac{x}{n}\right) \, . $$ which can be deduced geometrically:

Graph of the hyperbola

$\log\left(1+\frac{x}{n}\right)$ is the area of the region bounded by the hyperbola $y=1/t$, the $t$-axis, and the vertical lines $t=1$ and $t=1+\frac{x}{n}$. The rectangle with a width of $x/n$ and a height of $1$ gives us an upper bound for this region. A little care is needed when $-1 < x/n < 0$, but the result still holds. Hence, $$ n\log\left(1+\frac{x}{n}\right) \leq x \, , $$ and therefore $$ \exp\left(n\log\left(1+\frac{x}{n}\right)\right)=\left(1+\frac{x}{n}\right)^n \leq \exp(x) \, . $$ Note however that this proof is only valid for $\frac{x}{n}>-1$. If $\frac{x}{n} \le -1$, then $(1+\frac{x}{n})$ is nonpositive, and so the definition $\exp\left(n\log\left(1+\frac{x}{n}\right)\right)=\left(1+\frac{x}{n}\right)^n$ no longer applies.

Joe
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Work with the logarithm of both terms ...

Set $h(x) = x - n\log(1 + x/n)$ on $0 \leq x < \infty$. You have $h(0) = 0$ and \begin{align} h'(x) &= 1 - 1/(1 + x/n) \\ &= \frac{x/n}{1 + x/n}, \end{align} which is nonnegative, so $h$ is increasing for $x > 0$, that is, $h(x) \geq 0$.

A rural reader
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    Actually, we can state that $h(x)\ge0$ for $x>-n$, which is the domain of the function, because $0$ is a point of minimum. – egreg Jun 29 '21 at 21:53