I'm working through some of the AM-GM problems in Larson's Problem-Solving Through Problems and I'm stuck on $7.2.11.b$. Part $a$ was proving the below
Let $x_i \in \mathbb{R}^+$ and $p_i \in \mathbb{Z}^+$, then prove
$$(x_1^{P_1}\cdots x_n^{P_n})^{1/p_1+\cdots+p_n} \leq \frac{p_1x_1 + \cdots + p_nx_n}{p_1 + \cdots p_n}.$$
That is a straightfoward application of AM-GM. Part $b$ however says to show this inequality holds when $p_i \in \mathbb{Q}^+$.
Let $p_i = \frac{a_i}{b_i}$ and $y_i = x_i^{1/b_i}$ then
$$(x_1^{P_1}\cdots x_n^{P_n})^{1/p_1+\cdots+p_n} = (y_1^{a_1}\cdots y_n^{a_n})^{1/p_1+\cdots+p_n}.$$
And since $p_1 + \cdots p_n \leq a_1 + \cdots a_n$, we can extend the inequality,
$$(y_1^{a_1}\cdots y_n^{a_n})^{1/p_1+\cdots+p_n} \leq (y_1^{a_1}\cdots y_n^{a_n})^{1/a_1+\cdots+a_n}.$$
Now we apply AM-GM and get that
$$(y_1^{a_1}\cdots y_n^{a_n})^{1/a_1+\cdots+a_n} \leq \frac{a_1y_1 + \cdots + a_ny_n}{a_1 + \cdots a_n} \leq \frac{a_1y_1 + \cdots + a_ny_n}{p_1 + \cdots p_n}.$$
And this is where I'm stuck. If $a_iy_i \leq p_ix_i$ then we're done. But I can't convince myself of that since $x$ can be in $(0,1)$. Can anyone nudge me in the right direction?
We may assume, by using a common denominator for all $p_i$'s, that $p_i = a_i/b$. Then we have \begin{align} (x_1^{p_1}\ldots x_n^{p_n})^{1/(p_1+\ldots p_n)} &= (x_1^{a_1/b}\ldots x_n^{a_n/b})^{1/(a_1/b + \ldots + a_n/b)} \\ &= (x_1^{a_1}\ldots x_n^{a_n})^{1/(a_1 + \ldots + a_n)} \\ &\leq \frac{a_1x_1 + \ldots a_nx_n}{a_1 + \ldots a_n}\\ &= \frac{(a_1/b)x_1 + \ldots (a_n/b)x_n}{a_1/b + \ldots a_n/b}\\ &= \frac{p_1x_1 + \ldots p_nx_n}{p_1 + \ldots p_n}, \end{align} where we have used the weighted AM-GM for positive integers in the middle line.
Old Answer:
If any of the $x_i$'s are $0$, we have equality. Otherwise, since the inequality is homogenous, it suffices to prove it for $\lambda x_i$, for some $\lambda$. Choose $\lambda$ large enough so that $(\lambda x_i)^{1/b_i} < \lambda x_i / b_i$ for all $x_i$. Then we have $a_i(\lambda x_i)^{1/b_i} \leq p_i (\lambda x_i)$.