Generalizing Titu's Lemma

Question

I found a nice generalization of Titu's Lemma and was wondering if this has a name or has a reference anywhere.

Let $m$ be an integer greater than or equal to 2, $a_i^m$ a non-negative real number, and $x_i$ a positive real number. Then,

$$n^{m-2} \sum_{i=1}^{n}\frac{a_i^m}{x_i} \geq \frac{\left(\sum_{i=1}^{n}a_i\right)^m}{\sum_{i=1}^{n}x_i}$$

This reduces to Titu's Lemma when $m = 2$

Proof: By Hölder's inequality,

$$\left(\sum_{i=1}^{n}1\right)^{\frac{m-2}{m}}\left(\sum_{i=1}^{n}\frac{a_i^m}{x_i}\right)^{\frac{1}{m}} \left(\sum_{i=1}^{n}x_i\right)^{\frac{1}{m}} \geq \sum_{i=1}^{n}1^{\frac{m-2}{m}}\left(\frac{a_i^m}{x_i}\right)^\frac{1}{m}x_i^\frac{1}{m}$$

$$n^{\frac{m-2}{m}}\left(\sum_{i=1}^{n}\frac{a_i^m}{x_i}\right)^{\frac{1}{m}} \left(\sum_{i=1}^{n}x_i\right)^{\frac{1}{m}} \geq \sum_{i=1}^{n}a_i$$

$$n^{m-2} \sum_{i=1}^{n}\frac{a_i^m}{x_i} \geq \frac{\left(\sum_{i=1}^{n}a_i\right)^m}{\sum_{i=1}^{n}x_i}$$

Answer 1

For non-negatives $a_i$ it's just Holder for three sequences: $$n^{m-2}\sum_{i=1}^n\frac{a_i^m}{x_i}=\frac{\left(\sum\limits_{i=1}^n1\right)^{m-2}\sum\limits_{i=1}^n\frac{a_i^m}{x_i}\sum\limits_{i=1}^nx_i}{\sum\limits_{i=1}^nx_i}\geq$$ $$\geq\frac{\left(\sum\limits_{i=1}^n\left(1^{m-2}\cdot \frac{a_i^m}{x_i}\cdot x_i\right)^{\frac{1}{m-2+1+1}}\right)^{m-2+1+1}}{\sum\limits_{i=1}^nx_i}=\frac{\left(\sum\limits_{i=1}^na_i\right)^m}{\sum\limits_{i=1}^nx_i}.$$ For example, for positives $a$, $b$ and $c$ we have: $$\sum_{cyc}\frac{a}{b+c}=\sum_{cyc}\frac{a^3}{a^2(b+c)}\geq\frac{(a+b+c)^3}{3\sum\limits_{cyc}(a^2b+a^2c)}.$$

Answer 2

I found an even more general form in the paper Generalizations and Refinements for Bergström and Radon's Inequalities The paper refers to it as a generalization of Radon's inequality:

If $a_k, x_k > 0$, $p>0$, and $r \geq p+1$, then

$$\sum_{k=1}^n\frac{x_k^r}{a_k^p} \geq \frac{1}{n^{r-p-1}} \cdot \frac{\left(\sum_{k=1}^n x_k\right)^r}{\left(\sum_{k=1}^n a_k\right)^p}$$

When $p = 1$ and $r = 2$, it becomes Titu's Lemma (also known as Bergström's inequality or Sedrakyan's inequality).

When $r = p+1$, it becomes Radon's inequality.

When $p=1$, it becomes the form posed in this question.