Prove $(1+\frac{1}{x})^x(1+\frac{1}{x})>e$

Question

Show that $(1+\frac{1}{x})^x(1+\frac{1}{x})>e$ for $x>0$. My idea is:
$(1+\frac{1}{x})^x(1+\frac{1}{x})>(1+\frac{1}{x})^x\rightarrow e$, as $x\rightarrow \infty $. So from the permanence theorem of the sign I conclude that $(1+\frac{1}{x})^x(1+\frac{1}{x})>e$.
to do: check my idea

Answer 1

It doesn't work, because $$ \Bigl(1+\dfrac{1}{x}\Bigr)^{\!x}<e $$ You want to prove that, for $x>0$, $$ \Bigl(1+\dfrac{1}{x}\Bigr)^{\!x+1}>e $$ which is equivalent to $$ f(x)=(x+1)\log\Bigl(1+\dfrac{1}{x}\Bigr)>1 \tag{*} $$ Consider $$ f'(x)=\log\Bigl(1+\dfrac{1}{x}\Bigr)+(x+1)\Bigl(\frac{1}{x+1}-\frac{1}{x}\Bigr) =1-\Bigl(1+\frac{1}{x}\Bigr)+\log\Bigl(1+\dfrac{1}{x}\Bigr) $$ OK, let's simplify the thing: set $g(t)=1-t+\log t$, for $t>1$; then $g'(t)=(1-t)/t$<0. Hence $g$ is decreasing and its limit at $1$ is $0$. Thus $g(t)<0$ for $t>1$ and we conclude that $f'(x)<0$.

Therefore $f$ is decreasing; since its limit at $\infty$ is $1$, we have verified the inequality (*).

We can also verify the first statement in a similar way. Consider $$ F(x)=x\log\Bigl(1+\frac{1}{x}\Bigr) $$ Then $$ F'(x)=\log\Bigl(1+\frac{1}{x}\Bigr)+x\Bigl(\frac{1}{x+1}-\frac{1}{x}\Bigr)=-\log\Bigl(\frac{x}{x+1}\Bigr)+\frac{x}{x+1}-1 $$ Consider $G(t)=t-1-\log t$, so that $G'(t)=(t-1)/t$, which is negative for $0<t<1$ and $G(1)=0$. Therefore $F'(x)>0$ and so $F$ is increasing. Since its limit at $\infty$ is $1$, we have the desired inequality.

Answer 2

Similar to @egreg's answer, consider that you look for the minimum value of $$f(x)=(x+1)\log\Bigl(1+\dfrac{1}{x}\Bigr) $$ We have $$f'(x)=\log \left(1+\frac{1}{x}\right)-\frac{1}{x}\qquad \text{and} \qquad f''(x)=\frac{1}{x^2(x+1)}$$ The first derivative is always negative and the second derivative is positive.

Expanding $f(x)$ as a series $$f(x)=1+\frac{1}{2 x}-\frac{1}{6 x^2}+\frac{1}{12 x^3}+O\left(\frac{1}{x^4}\right)$$ $$e^{f(x)}=e\left(1+\frac{1}{2 x}-\frac{1}{24 x^2}+\frac{1}{48 x^3}+O\left(\frac{1}{x^4}\right)\right)$$