# Prove $-a\cdot e\cdot \ln(x)\le x^{-a}$

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?

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

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