For which positive numbers $a$ is it true that $a^x \ge 1 + x$ for all $x$? For which positive numbers $a$ is it true that $a^x \ge 1 + x$ for all $x$?
This is question 24, chapter 3 from Stewart's Early Transcendental problems plus.
I think I'm on the right track, but I'm not sure how to prove it rigorously.
I simplified the inequality to $a \ge (1 + x)^\frac{1}{x}$. Thus, the inequality is true for all values of $a$ that are greater than the function $(1 + x)^\frac{1}{x}$ for all values of $x$.
I'm aware that $\lim_{x \to \infty}(1 + x)^\frac{1}{x} = e$, and so therefore I assume that the the inequality is true for all values where $a \ge e$. However, I need to show that $(1 + x)^\frac{1}{x}$ is an increasing function to do this, and that seems a bit complicated.
Am I on the right track? Or is there an easier way to solve this.
Any help is appreciated.
Edit: I forgot to mention, but: if I do need to prove this function is increasing, I guess I just need to show that there is no value for which the function has a negative slope. Taking the derivative, this means I have to prove that $\frac{x}{x + 1} \ge \ln(x + 1)$ for all positive values of $x$, but this seems more complicated than it should be.
 A: Let
$$f(x)=\ln\left[(1+x)^{1/x}\right]=\frac 1x\ln(1+x)=\frac 1x\int_1^{1+x}\frac{dt}t$$
for $x>0$.
Let's prove that $f$ is decreasing. Take any $x>0$ and $y>x$.
Then 
$$\begin{align}
f(y)-f(x)&=\frac 1y\int_1^{1+y}\frac{dt}t-\frac 1x\int_1^{1+x}\frac{dt}t\\
&\phantom{m}\\
&=\frac1y\int_{1+x}^{1+y}\frac{dt}t-\left(\frac 1x-\frac 1y\right)\int_1^{1+x}\frac{dt}t\\
&\phantom{m}\\
&\leq\frac{y-x}{y(x+1)}-\frac{y-x}{xy}\ln(1+x)\\
&\phantom{m}\\
&= \frac{y-x}y\left[\frac 1{x+1}-\frac {\ln(1+x)}x\right]\\
&\phantom{m}\\
&=\frac{y-x}{xy(x+1)}[x-(x+1)\ln(x+1)]
\end{align}$$
To finish we need to show that $x-(x+1)\ln(x+1)<0$. For that, define
$$g(x)=x-(x+1)\ln(x+1)$$
$$g'(x)=-\ln(x+1)<0$$
Since $g(0)=0$, $g$ is negative for $x>0$ and $f$ is decreasing.
Now, we need to compute $\lim_{x\to 0^-}f(x)$. We apply l'Hopital's rule:
$$\lim_{x\to 0^-}f(x)=\lim_{x\to 0^-}\frac{1/(1+x)}{1}=1$$
That means that $f(x)<1$ for $x>0$ and
$$(1+x)^{1/x}<e$$
A: First, prove that $1+x\le e^x$. Then, the limit tells us that we can an $x$ sufficiently close to $0$ such that $\sqrt[x]{1+x}=e-\epsilon$ for every $\epsilon >0$. So if $a<e$, then we can find an $x$ such that $1+x> a^x$
