3
$\begingroup$

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.

$\endgroup$
1

2 Answers 2

3
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

By Holder's inequality: $$\left(\sum \frac{x^{p+1}}{y^p}\right)^{\frac{1}{p+1}} \left(\sum y\right)^{\frac{p}{p+1}} \ge \sum x \tag*{$\blacksquare$}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .