proof of following inequality

Question

could some one prove the following inequality,I think it is based on Taylor's expansion but I don't how to start

$$1-x \geq e^{-x-x^2} $$

and also for what value of x does it hold ?

another question is that I know that $1-x \geq e^{-x}$, so is the following also true $1-x \geq e^{-x-x^2-x^3}$ so how do we prove such inequalities?

Answer 1

$y=1-x$ is the equation of the tangent at $(0,1)$. Now the function $f(x)=\mathrm e^{-x-x^2}$ is concave in a neighbourhood of $x=0$: $$f'(x)=(-1-2x)e^{-x-x^2},\quad f''(x)=\bigl(-2+(-1-2x)^2\bigr)e^{-x-x^2}=(4x^2+4x-1)e^{-x-x^2},$$ so $f''(x)<0$ between the roots of $4x^2+4x-1$, $\;\frac{1\pm\sqrt 2}2$.

Therefore $\;1-x>e^{-x-x^2}$ for all $x\in\bigl(\frac{1-\sqrt 2}2,0\bigr)\cup\bigl(0,\frac{1+\sqrt 2}2\bigr) $.

Answer 2

Both inequalities they are wrong. Try $x=1$.

Answer 3

Equality leads to transcendental equations, which will require numerical solutions.