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Today during a Calc I exam we(me and my classmates) have been ask to prove $$-a\cdot e\cdot \ln(x)\le x^{-a}$$, $\forall x>0$ and $\forall a\in \mathbb{R}$. But, noneone in the room has known how to do it. Anyone knows how to?

Thanks in advance

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  • $\begingroup$ $-a \cdot \ln(x) = \ln(x^{-a})$. Then exponentiate both sides. $\endgroup$ Jan 22, 2021 at 20:24

2 Answers 2

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HINT

Try to find the maximum of $f(t):=\frac{\ln(t)}{t}$ and recall that your inequality is equivalent to $$ \frac{\ln(x^{-a})}{x^{-a}}\le\frac{1}{e}.$$

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The inequality is obviously true if $a=0$. If $a\not=0$, let $x=e^{-u/a}$. The inequality to prove becomes $eu\le e^u$ for all $u\in\mathbb{R}$. Now $f(u)=e^u-eu$ has a unique critical point at $u=1$, that being where $f'(u)=e^u-e=0$. Since $f''(u)=e^u\gt0$ for all $u$, the critical point at $u=1$ is a global minimum, and thus

$$e^u-eu=f(u)\ge f(1)=e^1-e\cdot1=0$$

for all $u$, which proves the desired inequality.

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