Proof of inequality $\frac{2-a}{2+a}<e^{-a}$

Question

How can I prove that

$$\frac{2-a}{2+a}<e^{-a}$$

for all $a \geq 0$ ?

For $a \geq 2$ it is clear, but how can it be shown for $0<a<2$ ?

Answer 1

Consider $f(x) = e^{-x}(2+x)$.

$$ f'(x) = -e^{-x}(1+x) \\ f''(x) = xe^{-x} > 0 $$ hence $f$ is convex on the domain $\{x:x>0\}$.

The linear approximation around $0$ of $f$ is: $$ g(x) = f(0) + f'(0) x = 2 - x $$hence $$x>0\implies f(x) > g(x) \\ \implies e^{-x} > \frac{2-x}{2+x} $$

Answer 2

The inequality is false for $a\in[-2,0]$, as you can easily check. I have included both endpoints because the lhs is undefined at $-2$ and they are equal at $0$.

To prove it for $a>0$ the easiest way is just to use the series, lhs is $1-a+\frac{a^2}{2}-\frac{a^4}{4}+\frac{a^4}{8}+\dots$, but rhs is $1-a+\frac{a^2}{2}-\frac{a^4}{6}+\frac{a^{24}}{8}+\dots$

A trivial induction on the coefficients establishes the result.

[Added later] The question has now been changed to require $a\ge 0$. The result remains false at $a=0$.

[Added still later] Of course, more elementary approaches are possible, like mathlove's solution. But the question is tagged calculus, so presumably one is allowed to use it.]

Answer 3

$$\frac{2-a}{2+a}\lt e^{-a}\tag1$$ We know that $(1)$ holds for $a\lt -2\ \text{or}\ a\ge 2$ trivially.

So, let us consider the case when $0\lt a\lt 2$ (Note that $(1)$ does not hold for $a=0$).

Since $$(1)\iff f(a)=a+2+(a+2)e^a-4e^a\gt 0,$$ we have $$f'(a)=e^a(a-1)+1,\ \ f''(a)=ae^a\gt 0.$$ Since $f'(a)$ is increasing with $f'(0)=0$, we know that $f'(a)\gt 0$ for $0\lt a\lt 2$. Since $f(a)$ is increasing for $0\lt a\lt 2$, we have $f(a)\gt f(0)=2+2-4=0$ for $0\lt a\lt 2$. Q.E.D.

Answer 4

Let $$f(a)=\frac{2-a}{2+a}-e^{-a}$$ Now $f'(a)=e^{-a}-\frac4{(a+2)^2}$ $f'(a)<0\;\forall\; a\ge0$ and also $f(0)=f'(0)=0$ You can do that probably by differentiating further.