Assume by contradiction that
$$\frac{x_1}{y_1^p} + \frac{x_2}{y_2^p} + \dots + \frac{x_n}{y_n^p} < x_1 + x_2 + \dots + x_n.$$
On the other hand, by multiplying the given relation by $p \in \mathbb{N}$, we have
$$p(x_1 + x_2 + \dots + x_n) \geq p(x_1 y_1 + x_2 y_2 + \dots + x_n y_n).$$
Adding the above inequalities we get
$$(p+1) \sum_{i=1}^{n} x_i > \sum_{i=1}^{n} x_i \left( p y_i + \frac{1}{y_i^p} \right).$$
But by the AM-GM inequality we have
$$p y_i + \frac{1}{y_i^p} \geq p + 1.$$
Finally, $(p+1) \sum_{i=1}^{n} x_i > (p+1) \sum_{i=1}^{n} x_i$, which is false.

We shall use the following inequality due to Radon.

LEMMA. If $x_1, x_2, \dots, x_n$ and $y_1, y_2, \dots, y_n$ are positive real numbers, and $p$ is a non-negative real number, then the following inequality holds
$$\frac{x_1^{p+1}}{y_1^p} + \frac{x_2^{p+1}}{y_2^p} + \dots + \frac{x_n^{p+1}}{y_n^p} \ge \frac{(x_1 + x_2 + \dots + x_n)^{p+1}}{(y_1 + y_2 + \dots + y_n)^p}.$$
Proof. We can rewrite the inequality as
$$\sum_{i=1}^{n} y_i \left( \frac{x_i}{y_i} \right)^{p+1} \ge \left( \sum_{j=1}^{n} y_j \right) \left( \frac{\sum_{i=1}^{n} x_i}{\sum_{i=1}^{n} y_i} \right)^{p+1}$$
or equivalently
$$\sum_{i=1}^{n} \frac{y_i}{\sum_{j=1}^{n} y_j} \left( \frac{x_i}{y_i} \right)^{p+1} \ge \left( \frac{\sum_{i=1}^{n} x_i}{\sum_{i=1}^{n} y_i} \right)^{p+1}.$$
Let us consider the convex function $f : [0, \infty) \to \mathbb{R}$ defined by $f(x) = x^{p+1}$. The last inequality follows from Jensen's inequality
$$\lambda_1 f(\alpha_1) + \lambda_2 f(\alpha_2) + \dots + \lambda_n f(\alpha_n) \ge f(\lambda_1 \alpha_1 + \dots + \lambda_n \alpha_n),$$
where $\lambda_i = \frac{y_i}{y_1 + y_2 + \dots + y_n} > 0, \sum_{i=1}^{n} \lambda_i = 1$ and $\alpha_i = \frac{x_i}{y_i}$. $\square$

Now, returning to our problem, by applying the LEMMA and the hypothesis of the problem, we obtain
$$\sum_{i=1}^{n} \frac{x_i}{y_i^p} = \sum_{i=1}^{n} \frac{x_i^{p+1}}{(x_i y_i)^p} \ge \frac{\left(\sum_{i=1}^{n} x_i\right)^{p+1}}{\left(\sum_{i=1}^{n} x_i y_i\right)^p} \ge \sum_{i=1}^{n} x_i.$$