An exponential/polynomial inequality

Question

Prove that there is at least $1$ real number $a>0$ with the property $$a^x\ge x^a $$ for any $x>0$.

Answer 1

I would do it this way:

Let $f(x)=\dfrac{x}{\log x}$ for some $x>0$. Hence, $f^\prime(x)=\dfrac{\log x-1}{\log^2 x}$. I can conclude that $f(x)\geq e$ $\forall\, x>0$.

Therefore, $$f(x)=\dfrac{x}{\log x}\geq e,$$ so, $$x\geq \log x^e,$$ hence, $$e^x\geq x^e.$$

Finally, $\exists a\,>0$ where $a=e$ and $a^x\geq x^a$

Answer 2

xyz's method has the right idea. If you consider the equation $$a^x \ge x^a$$ we can rewrite this as the equivalent $$\ln(a)/a \ge \ln(x)/x$$ by using the properities of the logarithm and the positivity of $x$ and $a$.

Thus if you find the $x$ that maximizes $\ln(x)/x$, that is the $a$ you are looking for.