Proving $x-x\ln x\leq e^{-x}+x^2$, for $x>0$ 
When $x>0$, prove :
$$x-x\ln x\leq e^{-x}+x^2$$

I tried: Let
$$f(x)=e^{-x}+x^2-x+x\ln x$$
Then
$$f'(x)=-e^{-x}+2x+\ln x$$
It looks like useless for the proof.
Who can help me? thks!
 A: Let $f(x) = \mathrm{e}^{-x} + x^2 + x\ln x - x$. We have
$$f'(x) = - \mathrm{e}^{-x} + 2x + \ln x$$
and
$$f''(x) = \mathrm{e}^{-x} + 2 + \frac{1}{x}.$$
Since $f''(x) > 0$ for all $x > 0$, $f'(x)$ is strictly increasing on $x > 0$.
Since $f'(\mathrm{e}^{-1}) \le 2\mathrm{e}^{-1} + \ln (\mathrm{e}^{-1}) < 0$
and $f'(1) > 0$, there exists a unique $x_0\in (\mathrm{e}^{-1}, 1)$ such that $f'(x_0) = 0$
that is
$$\mathrm{e}^{-x_0} - x_0 = x_0 + \ln x_0. \tag{1}$$
Also $f'(x) > 0$ on $(x_0, \infty)$ and $f'(x) < 0$ on $(0, x_0)$. Thus, we have $f(x) \ge f(x_0)$ for all $x > 0$.
(Note: Actually, $f(x)$ is convex on $x > 0$.)
We claim that $x_0 + \ln x_0 \ge 0$. Indeed, if $x_0 + \ln x_0 < 0$,
then we have $\ln x_0 < - x_0$ and thus $x_0 < \mathrm{e}^{-x_0}$
and thus $\mathrm{e}^{-x_0} - x_0 > 0$ which contradicts (1).
By using (1), we have
$$f(x_0) =  (2x_0 + \ln x_0) + x_0^2 + x_0\ln x_0 - x_0 = (1 + x_0)(x_0 + \ln x_0) \ge 0.$$
Thus, $f(x) \ge 0$ for all $x > 0$.
We are done.
A: Try and evaluate both sides at zero:
$$f(x)=x\left[1-\ln(x)\right]$$
$$f(0)=0$$
$$g(x)=e^{-x}+x^2$$
$$g(0)=1$$
so it is clear that this expression hold for $0$, now I would suggest looking at $f(x)$.
$$f(x)=x\left[1-\ln(x)\right]=0\Rightarrow x=0,e$$
so we know that at $x=e$, $LHS=0$ and from calculating the derivative:
$$f'(x)=-\ln(x)$$ it is clear that this function is decreasing where $\ln(x)>0\Rightarrow x>1$, and since it is clear that $g'(x)>0\forall x>0$ it is clear that $f(x)<g(x)\forall x>e$. Now we just need to show that $f'(x)<g'(x)\forall(0<x<e)$
