## Monday, August 5, 2013

### 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\$