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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.

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4 Answers 4

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$$ x^{-a}+ae \ln(x) = e^{-a \ln(x)} + ae \ln(x) $$ 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: $$ e^u = f(u) \ge f(1) + (u-1)f'(1) = e + (u-1)e = eu \, . $$

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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.

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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 $$f_x(-1/\ln x)=0.$$

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    $\begingroup$ Nice proof (+1) $\endgroup$
    – Angelo
    Feb 4, 2021 at 19:02
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Hint:
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 $x\in (0,+\infty)$

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