# Prove that $\sum_{i=1}^n \frac{x_i^{p+1}}{y_i^p} \ge \frac{\left(\sum_{i=1}^n x_i\right)^{p+1}}{\left(\sum_{i=1}^n y_i\right)^p}$

Prove that for positive real numbers $$p,x_1,\cdots, x_n, y_1,\cdots, y_n,$$ we have $$\sum_{i=1}^n \frac{x_i^{p+1}}{y_i^p} \ge \frac{\left(\sum_{i=1}^n x_i\right)^{p+1}}{\left(\sum_{i=1}^n y_i\right)^p}$$

By the power-mean inequality, we have that for positive real numbers $$x_1,\cdots, x_n$$ and numbers $$k\ge m$$, $$\left( \frac{\sum_{i=1}^n x_i^{k}}{n}\right)^{1/k} \ge \left(\frac{ \sum_{i=1}^nx_i^{m}}{n}\right)^{1/m}.$$ Holder's inequality says that for real numbers $$x_1,\cdots, x_n, y_1,\cdots, y_n,$$ we have $$\sum_{k=1}^n |x_k y_k| \leq (\sum_{k=1}^n |x_k|^p)^{1/p} (\sum_{k=1}^n |y_k|^q)^{1/q}$$, where $$p,q \in (1,\infty), 1/p + 1/q = 1.$$ There's also the Cauchy-Schwarz inequality, which solves the problem in the special case where $$p=1$$. I'm not sure if Jensen's inequality might be useful for the general case.

We can prove this inequality for the case $$p \ge 1$$ using the classical Jensen inequality. First observe that by dividing both sides by the right side and rearrange the factors, we can further assume that $$\displaystyle \sum_{i=1}^n x_i = 1 = \displaystyle \sum_{i=1}^n y_i$$. And also let $$u_i = \dfrac{x_i}{y_i}$$. Then the inequality becomes: $$\displaystyle \sum_{i=1}^n y_iu_i\cdot u_i^p\ge 1$$, or $$\displaystyle \sum_{i=1}^n y_iu_i^{p+1}\ge 1$$. Then since $$p+1 \ge 2$$, and the function $$f(u) = u^{p+1}$$ is convex on $$(0,\infty)$$, then by Jensen inequality: $$\displaystyle \sum_{i=1}^n y_iu_i^{p+1}\ge \left(\displaystyle \sum_{i=1}^n y_iu_i\right)^{p+1}= \left(\displaystyle \sum_{i=1}^n x_i\right)^{p+1}=1^{p+1} = 1.$$. We arrived at the desired inequality.
By Holder's inequality: $$\left(\sum \frac{x^{p+1}}{y^p}\right)^{\frac{1}{p+1}} \left(\sum y\right)^{\frac{p}{p+1}} \ge \sum x \tag*{\blacksquare}$$