Fractional inequality

Let $x, y > 0$, $x \neq y$, and let $n > 1$ be an integer.

If $x^n - y^n = x^{n+1} - y^{n+1}$, show that $$1 < x+y < \frac{2n}{n+1}$$

Solution Let $f(x) = x^n ( 1 - x)$ and we have $f(x) = f(y)$. WLOG, we may assume that $x If $y > 1$ then $f(y)$ is negative, so that $f(x)$ is negative, which means $x > 1$ as well. But $f(x)$ is monotonically decreasing for $x > 1$ (because $x^n$ is monotonically increasing and $1-x$ is monotonically decreasing). Contradiction.

If $y=1$ then $f(x) = f(y) = 0$ impossible for $x>0$. Contradiction.

If $0 < x < y < 1$, we draw the graph of $f(x)$. It starts off at (0,0), going up to a maximum between 0 and 1, and goes back down to (1,0). This graph can be deduced easily be realizing that $f(x) = x(1-x) . x^{n-1}$. Because $x(1-x)$ only has one turning point and $x^{n-1}$ is monotonically increasing (or constant), then $f(x)$ also only has one turning point.

We can find that turning point by using AM-GM: $$f(x) = x.x\dots x.(1-x)$$ $$ = \frac{1}{n} x.x\dots x.(n-nx)$$ $$ \leq \frac{1}{n} (\frac{n}{n})^n = \frac{1}{n}$$ Equality happens when $x = n-nx$ or $x = \frac{n}{n+1}$. Note that for $n>1, \frac{n}{n+1} = 1 - \frac{1}{n+1}> 1/2$