# 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: $$e^u = f(u) \ge f(1) + (u-1)f'(1) = e + (u-1)e = eu \, .$$
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.
Let $$f_x(a):=e^{-a\ln x}+ae\ln x.$$ For $$x=1$$, it is clear that $$f_1(a)=1\geq0$$. Now let $$x\neq0$$, then $$f'_x(a)=-e^{-a\ln x}\ln x +e\ln x\overset{!}{=}0\Leftrightarrow a=-1/\ln x.$$ Now you just have to think about, why this is a minimum point and thus the minimum of the function is $$f_x(-1/\ln x)=0.$$
Define the function $$f:(0,+\infty)\to\Bbb R$$ by $$f(x)=x^{-a}+ae\ln(x)$$ Show that $$f(x)\to+\infty$$ as $$x\to0^+$$ and $$x\to +\infty$$. Conclude that $$f$$ attains its global minimum. Use its derivative to find where it is. You will see that $$f(x)=0$$ at that point, so we get $$f(x)\geq0$$ for all $$x\in (0,+\infty)$$