If x and y are two consecutive positive integer then how can we prove that $y^y>x^x$?

Question

Actually we can prove it by just choosing random consecutive positive integers, but I want to know how we can prove it mathematically.

Answer 1

Suppose $y>x >0$.

Claim: $y^y > y^x$. $y^y - y^x = y^{y-x}(y^x - 1)$. Then $y^x > 1^x = 1$, so $(y^x -1) > 0$, so $y^{y-x}(y^x -1 ) > 0$. So the claim holds.

Claim: $y^x > x^x$. We have $y>x$, so $y^2 > yx > x^2$, proceeding inductively gives $y^x>x^x$.

Putting the two together gives the desired result.

Answer 2

Let $f(x) = x^x,$ then $f'(x) = f(x)\cdot(\log(x)+1).$

$\forall x \in \mathbb{Z}^+. f'(x) > 0 \implies \forall x \in \mathbb{Z}^+. f(x) \text{ is strictly increasing.}$