Show that $\left(1 + \frac{x}{n}\right)^{-n} \le 2^{-x}$, when $x,n \ge 0$, $x \le n$ Show that $\left(1 + \frac{x}{n}\right)^{-n} \le 2^{-x}$, when $x,n \ge 0$, $x \le n$. This is driving me crazy... I have plotted the graphs to be sure that the inequality is true, and it is, but I can't seem to show it. Here is what I have so far:
$$\left(1 + \frac{x}{n}\right)^{-n} \le \left(1 + \frac{x}{n}\right)^{-x}$$
since $n \ge x$. I want to then use the fact that $1 + \dfrac{x}{n} \le 2$, but that gives me a $\ge$, not a $\le$, so my first step must be wrong... I don't know what else to do, instead.
 A: Consider the function $$f(n,x)=\left(1+\frac{x}{n}\right)^{-n}-2^{-x}$$ What you can notice is that $$f(n,0)=0$$ $$f(n,n)=0$$ Computing the derivative $$\frac{df(n,x)}{dx}=2^{-x} \log (2)-\left(1+\frac{x}{n}\right)^{-n-1}$$ you also find $$\frac{df(n,0)}{dx}=\log (2)-1 <0$$ $$\frac{df(n,n)}{dx}=2^{-n} \log (2)-2^{-n-1}=2^{-n} \big(\log (2)-\frac 12\big)>0$$ So, it exists a value $a_n$ of $x$ such that $\frac{df(n,a_n)}{dx}=0$.
I think that the value of the function and its derivative at the bounds could be sufficient.
Edit
Just for your curiosity, there is an analytical expression for $a_n$; it is rather complex (involving Lambert function) and it is such that $$0\lt a_n \lt \frac{\log (\log (2))}{\log (2)-1}\approx 1.19443$$ that is to say in the range of interest.
Computing the second derivative of the function at this limit proves that it is always positive which implies that $a_n$ corresponds to a minimum.
A: First, this is equivalent to $$\left(1+\frac xn\right)^n \geq 2^{x}$$
Letting $y=\frac{x}{n}$, this is equivalent to, for $y\in[0,1]$ and any $n>0$:
$$(1+y)^n \geq 2^{yn}$$ or:
$$1+y\geq 2^y$$
Now we use that $f(x)=2^y$ is convex - that is, $f(at+b(1-t))\leq tf(a)+(1-t)f(b)$ for any $a,b$ in the domain of $f$ and $t\in[0,1]$. 
In this case, letting $a=1,b=0,t=y$, we'd have:
$$2^y=f(y)=f(1\cdot y + 0\cdot(1-y))\leq yf(1)+(1-y)f(0) = 1+y$$
A: I propose to see how do the limits of this expression compare to each other.
Recall the well-known limit:
$$
\lim_{k\to \infty} \left( 1+\frac{1}{k}\right)^{k} =
\lim_{k \to 0} \big( 1+k\big)^{\frac{1}{k}} =
e.
$$
Fix $x$ and let $n \to \infty$.
Denote  $m = \dfrac{n}{x}$, so that $x = \dfrac{n}{m} $,  $n = mx$, and $n\to \infty \iff m \to \infty $, then
$$
\left(1 + \frac{x}{n}\right)^{-n} \le \left(1 + \frac{x}{n}\right)^{-x} 
\stackrel{\text{ for fixed } x}{\iff}
\left(1 + \frac{1}{m}\right)^{-n} \le \left(1 + \frac{1}{m}\right)^{-\frac{n}{m}}
$$
Note that $x,n > 0$ and $x<n$ $\implies m > 1$.
Now, let us consider limits as $n\to \infty$ of left hand side and right hand side of inequality:

Let us start with the lhs of your inequality. 
$$
\lim_{n\to \infty} \left( 1+\frac{x}{n}\right)^{-n} = 
\lim_{n\to \infty} \bigg[\left( 1+\frac{x}{n}\right)^{n}\bigg]^{-1} = 
\lim_{m\to \infty} \bigg[\left( 1+\frac{1}{m}\right)^{mx}\bigg]^{-1} =
\bigg[\lim_{m\to \infty} \left( 1+\frac{1}{m}\right)^{m}\bigg]^{-x} =
e^{-x}
$$

Similarly, we do the same trick for the rhs of your inequality. 
$$
\lim_{n\to \infty} \left( 1+\frac{x}{n}\right)^{-x} = 
\lim_{n\to \infty} \bigg[\left( 1+\frac{x}{n}\right)^{\frac{n}{x}}\bigg]^{-\frac{x^2}{n}} = 
\lim_{m\to \infty} \bigg[\left( 1+\frac{1}{m}\right)^{m}\bigg]^{-\frac{x}{m}} =
\lim_{m\to \infty} e^{-\frac{x}{m}} =
\lim_{n\to \infty} e^{-\frac{x^2}{n}}  = 1
$$

Moreover, for the original inequality $\left(1 + \frac{x}{n}\right)^{-n} \le 2^{-x}$ we can use the limit for lhc, thus getting
$$
\boxed{e^{-x} \leq 2^{-x}} \iff
\left(\frac{e}{2}\right)^{-x} \leq 1\iff 
\left(\frac{2}{e}\right)^{x} \leq 1,
$$
which is always true for $x>0$, since $e>2 \implies \dfrac{2}{e} <1$.
A: I have already posted an answer where I considered asymptotic behavior of inequality as $n\to \infty$. 
Below are  more general conclusions.


Given $0\le x \le n$, show that $\left(1 + \dfrac{x}{n}\right)^{-n} \le 2^{-x}$.

First, note that 
$$
0\le x \le n \implies 0 \le \frac{x}{n}\le 1
\implies 1\le \left(1 + \dfrac{x}{n}\right) \le 2
$$
Second, note that for any $\alpha \ge 1$ we have
$$
\alpha \ge 1, \quad x \le n \implies 
\alpha^{x} \le \alpha^{n} \implies 
\alpha^{-x}  = \frac{1}{\alpha^{x}}\ge \frac{1}{\alpha^{n}} = \alpha^{-n}
$$
Third, for $1\le \alpha \le \beta$ and for $n \ge 0$ we have
$$
n\ge 0 \implies  \alpha \le \beta \iff\alpha^{n} \le \beta^{n} \iff 
\alpha^{-n} \ge \beta^{-n} 
$$
Fourth, rewrite $2^{-x}$ as $\left(2^{-\frac{x}{n}}\right)^{n}$, get
$$
2^{-x} = \left(2^{-\frac{x}{n}}\right)^{n}
\implies 
\left(1 + \dfrac{x}{n}\right)^{-n} \le 2^{-x} \iff 
\left(1 + \dfrac{x}{n}\right)^{-n} \le \Big(2^{\frac{x}{n}}\Big)^{-n}
$$
Finally, we only have to show that 
$ \left(1 + \dfrac{x}{n}\right) \ge 2^{\frac{x}{n}}$  whenever $0 \le x \le n$.
Let $y:= \frac{x}{n}\,\ln 2$, then
$$
\begin{aligned}
2^{\frac{x}{n}} = e^{\frac{x}{n}\cdot \ln 2}  = e^{y} 
& = \sum_{k=0}^{\infty}\frac{y^{k}}{k!}
= \sum_{k=0}^{\infty}\frac{\ln^{k}(2)}{n^{k}}\frac{x^{k}}{k!}
= 1 +  \frac{\ln2 }{n}x + \frac{\ln^{2}(2)}{n^{2}}\frac{x^{2}}{2} 
+ \frac{\ln^{3}(2)}{n^{3}}\frac{x^{3}}{6} + \cdots = \\
& = 1 +  \frac{\ln2 }{n}x + \mathcal{O}\left(x^2\right) 
\\
\left(1 + \dfrac{x}{n}\right)  = 1 + \dfrac{1}{n}x
& = 1 + \dfrac{1}{n}x + \mathcal{O}\left(x^2\right).
\end{aligned}
$$
Since $\ln 2< 1$, we conclude that $ 2^{\frac{x}{n}} \le \left(1 + \dfrac{x}{n}\right) $, and thus establish inequality:
$$
\begin{aligned}
\ln 2< 1 &\implies
1 + \dfrac{\ln 2}{n}x + \mathcal{O}\left(x^2\right) \le  1 + \dfrac{1}{n}x + \mathcal{O}\left(x^2\right)\implies 
\\ & \implies 
2^{\frac{x}{n}} \le \left(1 + \dfrac{x}{n}\right)   , \quad 0\le x \le n \implies 
\\ & \implies 
\boxed{\left(1 + \dfrac{x}{n}\right)^{-n} \le 2^{-x}}
\end{aligned}
$$
Q.E.D.
