Prove that for every $a \in ℝ$ and for every $x>0$ the inequality $x^{-a}+a · e · \ln(x) \geq 0$ holds I have tried some techniques to solve the following problem but I was unable to solve it:

"Prove that for every $a \in ℝ$ and for every $x>0$ the inequality $x^{-a}+a · e · \ln(x) \geq 0$ holds".

I tried to take the derivative and see if it is always positive, I tried a direct approach manipulating the expression, I also tried to divide it in cases, but I always find troubles.
I would appreciate any help or hint you could give me. Thanks.

Note that we can transform the inequality:
\begin{align}
x^{-a}+ae\ln(x) &\geq 0\\
\implies \ln\left(e^{x^{-a}}\right) + \ln\left(x^{ae}\right) &\geq 0\\
\implies \ln\left(e^{x^{-a}}x^{ae}\right) \geq 0\\
\implies e^{x^{-a}}x^{ae} \geq 1\\
\implies e^{x^{-a}} \geq x^{-ae}.
\end{align}
Now, setting $y = x^{-a}$, we note that $x>0, a\in \mathbb{R} \implies y > 0$, so we wish to prove that for all $y>0$ we have$$e^{y} \geq y^{e},$$ which is discussed here.
