Find $a\in (0,\infty)$ such that $a^x-x^a\geq0 \forall x>0$

Question

Find $a\in (0,\infty)$ such that $a^x-x^a\geq0 \forall x >0$

My attempt:

Let $f(x)=a^x-x^a$ then $f'(x)=a^x\ln a -ax^{a-1}$. $f'(x)=0\implies a^{x-1}\ln a=x^{a-1}$... and I can't really solve this equation. How should I approach this question differently?

Answer 1

Hint: Consider $g(x)=x^{1/x}$. Find its maximum.

Answer 2

Making $a = e$ we have

$$ e^{\frac{1}{e}} \ge x^{1/x} $$

Answer 3

Hint. Consider the function $h(x)=\frac{\ln(x)}{x}$ and note that for $x>0$, $$a^x\geq x^a\Leftrightarrow x\ln(a)\geq a\ln(x)\Leftrightarrow h(a)\geq h(x).$$ By evaluating the derivative of $h$, you should be able to find the point of global maximum $a$.

Answer 4

Let $f(x)=x\ln (a)-a \ln (x)$. If $a \geq 1$ then $f'(x)=\ln (a)-\frac a x$ which is 0 when $x=\frac a {\ln (a)}$. $f$ has minimum at this point so we need $a-a\ln {\frac a {\ln (a)}} \geq 0$. If $a<1$ then $f$ is decreasing tends to $-\infty$ so there is no $a$ for which $f$ is non-negative for all $x$. So the condition on $a$ is $a \geq 1$ and $\frac a {\ln (a)} \leq e$ or