How to show $\frac{2-x}{2+x} \le e^{-x}$ for $x\ge0$?

Question

Let $x\geq0$. Show that: $$\frac{2-x}{2+x} \le e^{-x}.$$

I have some troubles to prove it. Would you give me any hint?

Answer 1

Construct the function $f(x)=(2-x)-e^{-x}(2+x)$. We need to show that $f(x)\le 0,\ \forall x\ge 0$.

Now, note that $f'(x)=-1+e^{-x}(2+x)-e^{-x}=-1+e^{-x}(1+x)\le 0$ for $x\ge 0$. The last inequality follows since $e^{x}=\sum_{n\ge 0}x^n/n!\ge 1+x\ \forall\ x\ge 0$.

Thus, $f$ is a monotonically decreasing function for $x\ge 0$. Hence, $f(x)\le f(0)\ \forall x\ge 0$. Since $f(0)=0$, the result follows.$\quad\blacksquare$

Answer 2

We need to prove that $$(2-x)e^x\leq2+x$$ or $$x(e^x+1)\geq2(e^x-1)$$ or $f(x)\geq0$, where $$f(x)=x-\frac{2(e^x-1)}{e^x+1}$$ and calculate $f'(x)$: $$f'(x)=1-\frac{4e^x}{(e^x+1)^2}=\frac{(e^x-1)^2}{(e^x+1)^2}\geq0,$$ which gives $f(x)\geq f(0)=0$ and we are done!

Answer 3

We can assume that $0 \le x < 2$. Then, using the geometric series, $$ \frac{2+x}{2-x} = 1 + \frac{x}{1 - \frac 12 x} = 1 + \sum_{n=1}^\infty \frac{x^n}{2^{n-1}} \ge 1 + \sum_{n=1}^\infty \frac{x^n}{n!} = e^x \, . $$

Remark: $(2-x)/(2+x)$ is the Padé approximant of order [1/1] for $f(x) = \exp(-x)$, that is the (unique) rational function of the form $$ R(x) = \frac{a_0 + a_1 x}{1 + b_1 x} $$ such that $R(0) = f(0), R'(0) = f'(0)$ and $R''(0) = f''(0)$.