Show that $1 - x^2 \le e^{-x^2}$ for $0\le x\le1$ and $e^{-x^2}\le1/(1+x^2)$ for $0\le x$

Found this question in my freshman year calc course textbook under elementary integration. I used the following method:

$y=(1+t)e^{-t}$

$y'=-te^{-t}$

$y=1$ is maximum at $t=0$ taking $t=x^2$

$(1+x^2)e^{-x^2}\le1$

$e^{-x^2}\le1/(1+x^2)$

This is the second part of the question and I don't know how $x\ge0$ factors into this, shouldn't it work for all $x$? Any ideas on how I can prove the first part of the question?

This is what I have so far for the first part:

taking $t=-x^2$

$(1-x^2)e^{x^2}\le1$

$1-x^2 <= e^{-x^2}$

Again, don't know why it's only applicable for $0\le x\le1$

I suspect that the strange restrictions on $x$ are meant to encourage you to make the substitution $t=x^2$ and reframe the two exercises as

1. For $0\le t\le1$ show that $1-t\le e^{-t}$

2. For $t\ge0$ show that $e^{-t}\le\frac{1}{1+t}$

Then if you know the expansions

1. $e^{-t}=1-t+\dfrac{t^2}{2!}-\dfrac{t^3}{3!}+\cdots$

2. $e^t=1+t+\dfrac{t^2}{2!}+\dfrac{t^3}{3!}+\cdots$

you will be able to verify the inequalities.

For the first inequality, if $t\in[0,1]$, then for $k\ge2$, it is the case that $\dfrac{t^k}{k!}>\dfrac{t^{k+1}}{(k+1)!}$

So $1-t\le1-t+\dfrac{t^2}{2!}-\dfrac{t^3}{3!}+\cdots=e^{-t}$ [equal when $t=0$]

The second inequality is equivalent to $1+t\ge e^t$ which follows immediately from the expansion for $e^t$.