prove that $\frac{1-e^{-x^2}}{x}\le 2\sqrt{2} , \ x>0$,

Can you show very easy methods? I hope I'll see many methods. Thank you everyone.

Prove that:

$$\frac{1-e^{-x^2}}{x}\le 2\sqrt{2} \ \ \ \qquad \forall x>0.$$

Solution 1:

$$ 1-e^{-x^2}=\int_0^x2te^{-t^2}dt\;\Rightarrow\;\frac{1-e^{-x^2}}{x}=2\int_0^x\frac{t}{x}e^{-t^2}dt\leq 2 \int_0^xe^{-t^2}dt\leq 2\int_0^{+\infty}e^{-t^2}dt=\sqrt{\pi} $$

Solution 2:

We shall prove the stronger inequality $\frac{1-e^{-x^2}}{x} \leq \sqrt{e-1}<2\sqrt{2}$.

This is equivalent to $$1 \leq x\sqrt{e-1}+e^{-x^2}$$

If $x \geq \frac{1}{\sqrt{e-1}}$, then $$x\sqrt{e-1}+e^{-x^2} \geq 1+e^{-x^2} \geq 0$$.

Otherwise $x<\frac{1}{\sqrt{e-1}}$, then by Bernoulli's inequality $e^{-x^2} \geq 1-(e-1)x^2$, so $$x\sqrt{e-1}+e^{-x^2} \geq x\sqrt{e-1}+1-(e-1)x^2=1+x(\sqrt{e-1}-(e-1)x) \geq 1$$

where the last inequality holds since $\sqrt{e-1}-(e-1)x>\sqrt{e-1}-(e-1)\frac{1}{\sqrt{e-1}}=0$.

Solution 3:

It's clear for $x > 1$, the value is bounded above by $1$ since $1 - e^{-x^2} < 1$

If $0 < x \leq 1$ on the other hand, write

$$e^{-x^2} = 1 - x^2 + \dfrac{x^4}{2!} - \dots $$

and so $$\dfrac{1 - e^{-x^2}}{x} = x - \dfrac{x^3}{2!} + \dfrac{x^5}{3!} - \dots $$

This is an alternating series with decreasing coefficients since $x^n < x^m$ for $n > m$. Therefore it is bounded by the first term and so $$\dfrac{1 - e^{-x^2}}{x} < 1$$ for all $x$. (Thanks to @Andre for pointing out the simplification with the alternating series)