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.

$$
x^{-a}+ae \ln(x) = e^{-a \ln(x)} + ae \ln(x)
$$
so that, with the substitution $u = -a \ln (x)$, this is equivalent to showing that
$$
 e^u \ge eu
$$
for all $u \in \Bbb R$, and that is true because $f(u) = e^u$ is convex:
$$
 e^u = f(u) \ge f(1) + (u-1)f'(1) = e + (u-1)e = eu \, .
$$
