$x (1-e^{-x}) \leq 2 (e^{-x} + x - 1)$

Question

For $x \geq 0$ I want to prove the inequality $$ x (1-e^{-x}) \leq 2 (e^{-x} + x - 1) $$ using the fundamental theorem of calculus.

Let $f(x) = x (1-e^{-x})$. Then $f(0) = 0$ and $f'(x) = 1 - e^{-x} + x e^{-x}$. Hence $$ f(x) = \int_{0}^{x} (1 - e^{-y} + y e^{-y})dy = x + e^{-x} - 1 + \int_{0}^{x} y e^{-y}dy. $$ Now this already looks quite good and it remains to show that $$ \int_{0}^{x} y e^{-y}dy \leq x + e^{-x} - 1. $$ How can I do this?

Answer 1

We can use Question on the proof of $e^x>1+x$ for $x>0$

For $y\ge0$

$$\int_0^ye^tdt \ge\int_0^y dt$$

$$e^y-1\ge y$$

$$1-e^{-y}\ge ye^{-y}$$

Therefore,

$$\int_0^x ye^{-y}dy\le\int_0^x(1-e^{-y})dy=x+e^{-x}-1$$

Answer 2

Aliter approach

Let $f(x)=2(e^{-x}+x-1)-x(e^{-x}-1),$

$f(0)=0.$ We get $f'(x)>0.$ Hence, $f(x)$ is monotonic. So, $x\ge 0 \implies f(x)\ge f(0).$

Answer 3

Your inequality is equivalent to $$ e^{-x} \geq \frac{2-x}{2+x}\qquad\text{for }x\geq 0 \tag{1}$$ (which is trivial for $x\geq 2$) or to $$ e^{2x} \leq \frac{1+x}{1-x}\qquad\text{for }x\in[0,1) \tag{2}$$ or to $$ x \leq \text{arctanh}(x)\qquad\text{for }x\in[0,1) \tag{3}$$ which is trivial anyway, since $\text{arctanh}(x)$ is a convex function on $[0,1)$, or just by applying termwise integration to both sides of $$ 1 \leq \frac{1}{1-x^2}\qquad\text{for }x\in[0,1). \tag{4}$$