# Show that $(1+x)^{(1+x)}>e^x$

How can I prove that $$(1+x)^{(1+x)}>e^x$$ for all $$x>0$$?

The problem arose as I tried to prove the well-known & intuitive econometric principle that the more often you compound your interest, the more interest you ultimately get (in maths, that $$\frac{d}{dn}((1+\frac{1}{n})^n)>0$$ for $$n>0$$).

An interesting further problem is to prove that $$(a+x)^{(a+x)}>e^x$$ is true for all $$x>0$$ if and only if $$a\geq1$$.

• Take the logarithm of both sides, which results in $log(x+1)\gt \frac{x}{x+1}$. Then use the inequalities proven here: math.stackexchange.com/questions/324345/… – user665463 Oct 10 '19 at 12:33
• $\displaystyle ~(0.9+x)^{0.9+x} \gg e^x ~$ for $~x:=2~$ – user90369 Oct 16 '19 at 7:17

First, observe that \begin{align} &(1 + x)^{(1 + x)} > e^x \\&\iff (1 + x) \log (1 + x) > x \\&\iff (1 + x)\log(1 + x) - x > 0 \end{align} Set $$f(x) = (1 + x)\log(1 + x) - x$$. Then $$f'(x) = \log(1 + x) + 1 - 1 = \log(1 + x)$$ So $$f'(x) > 0$$ whenever $$x > 0$$. That is, $$f$$ is strictly increasing on $$(0, \infty)$$. On the other hand $$f(0) = (1 + 0) \log(1 + 0) - 0 = 0$$ These two facts together show that $$f(x) > 0$$ whenever $$x > 0$$, which is equivalent to what needed to be shown.

Hint: apply $$\ln$$ to both sides of the inequality.

$$(1+x)\ln(1+x) > x\Rightarrow (1+x)\ln(1+x) - x > 0$$

The proof for $$x \geq e-1$$ is straightforward. For $$0 < x < e-1$$, by the concept of derivative show that it is increasing in the specified area.

$$(1+x)^{1+x} = 1 + x + x^2 + \frac{x^3}{2} + \frac{x^4}{3} + \cdots$$ is clearly greater than $$e^{x} = 1 + x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{24} + \cdots$$ for $$x > 0.$$

Your general case is easy with this corollary of the Mean value theorem:

Let $$f,g$$ be two differentiable functions defined on an interval $$I$$ and $$x_0\in I$$.

If $$f(x_0)\ge g(x_0)$$ and $$f'(x)>g'(x)$$ for all $$x>x_0,\:x\in I$$, then $$f(x)>g(x)$$ for all $$x>x_0,\:x\in I$$.

Indeed it is enough to compare the logs: set $$f_a(x)=(a+x)\ln(a+x)$$, $$g(x)=x$$. We have $$f(0)=a\ln a\ge g(0)=0$$ iff $$a\ge 1$$.

Now compare the derivatives:

$$f'_a(x)=\ln(a+x)+1>1$$ for all $$x>0\iff a+x>1$$ for all $$x>0 \iff a\ge 1$$.

Most such inequalities seem to follow from the familiar result that $$e^t > 1 + t$$ for all real $$t \ne 0$$.

In this case, putting $$t = -\log(1 + x)$$: $$(1 + x)\log(1 + x) = e^{-t}(-t) > e^{-t}(1 - e^t) = e^{-t} - 1 = x \quad (x > -1,\ x \ne 0).$$