$\min_{ x>0}(T e^{-x}+ x^2)$ for all sufficiently large $T > 0$ Another real qual problem:

Show that for all sufficiently large $T > 0$, we have
$$0.9 \log^2 T\leq \min_{ x>0}(T e^{-x}+ x^2) \leq 1.1 \log^2 T$$

It seems a calculus problem, I started to take derivatives for $f(x)=T e^{-x}+ x^2$,$f'(x)=-Te^{-x}+2x$,$f''(x)=Te^{-x}+2$
but I did not reach to anything!! I'm Not sure if the approximation  $1+x\leq e^x$ play role in this problem or not!
Any help would be appreciated.
 A: Let $T = \mathrm{e}^S$.
It suffices to prove that, for all $S > 200$,
$$\frac{9}{10}S^2 \le \min_{x > 0} (\mathrm{e}^S \mathrm{e}^{-x} + x^2) \le \frac{11}{10}S^2.$$
We have, for all $S > 200$,
$$\min_{x > 0} (\mathrm{e}^S \mathrm{e}^{-x} + x^2) \le \mathrm{e}^S \mathrm{e}^{-S} + S^2 = 1 + S^2 < \frac{11}{10}S^2.$$
Thus, the right inequality is true.
For the left inequality:
Let
$$f(x) := \mathrm{e}^S \mathrm{e}^{-x} + x^2 - \frac{9}{10}S^2.$$
If $0 < x < S - 2\ln S$, we have
$$f(x) \ge \mathrm{e}^{2\ln S} - \frac{9}{10}S^2 =   \frac{1}{10}S^2 > 0.$$
If $x \ge S - 2\ln S$,
using $\mathrm{e}^u \ge 1 + u$ for all real numbers $u$, we have
$$\mathrm{e}^{S - x}
= S \mathrm{e}^{S - \ln S - x}
\ge S(1 + S - \ln S - x)$$
and for all $S > 200$,
\begin{align*}
 f(x) &\ge S(1 + S - \ln S - x) + x^2 - \frac{9}{10}S^2\\
 &= (x - S/2)^2 - S\ln S - \frac{3}{20}S^2 + S\\
 &\ge (S - 2\ln S - S/2)^2 - S\ln S - \frac{3}{20}S^2 + S\\
 &= \frac{1}{10}S^2 - 3S\ln S + 4\ln^2 S + S\\
 &> \frac{1}{10}S^2 - 3S\ln S\\
 &> 0
\end{align*}
where we have used $x - S/2 \ge S - 2\ln S - S/2 > 0$ for all $S > 200$.
We are done.
