Prove $x^{r}y^{1-r} ≤ rx + (1-r)y$ for $x,y > 0$ and $0<r≤1, r∈\mathbb{Q}$ using the AM-GM inequality

Question

Question is in the title.

I am stuck. This is what I've done so far:

$x^ry^{1-r} = x^ry^{-r}y = (x \cdot 1/y)^{r}y $

Using the AM-GM Inequality formula:

$ (x \cdot 1/y)^{r}y ≤ r(x + 1/y)y = rxy + r$

This is not the final result I want. Any ideas on how to get there?

Answer 1

You just write $r=\frac pq$ with natural $p,q$ and $p\leq q$.

Hence,

$$x^ry^{1-r} = \sqrt[q]{x^py^{q-p}}\leq \frac{px+(q-p)y}{q}$$