Checking inequality without actually calculating LHS and RHS

Question

How to check whether the following inequality is true or not without actually calculating the values of $x^y $ and $y^x $:

$$ x^y > y^x$$

(x and y are integers)

Answer 1

$x^y > y^x$ iff $y \ln x > x \ln y$ iff $\frac{\ln x}{x} > \frac{ \ln y} {y} $.

Consider the function $f(t) = \frac{ \ln t}{t} $.

$$ f' = \frac{1 - \ln t}{t^2} \implies f'(t) = 0 \; \; \; \text{iff} \; \; 1 = \ln t $$

Therefore, $\ max f = e $. In other words, $$x^y > y^x \; \; \; \text{Holds only if } x > y$$ and they both must be less $e$. $e$ is the exponential number. And both $x$ and $y$ must be positive.

And If both $x,y $ are greater than $e$, and $y > x$ then your inequality also holds.

Answer 2

If it were true then $x^y > y^x$ but swapping $x$ and $y$ gives $y^x > x^y$ which is a contradiction.