Suppose I have $a_1,\cdots,a_r$ with $a_i\geq 0, \forall i$.
How to prove $$\sum_{i=1}^r (r-1)a_i^2\geq\sum_{i,j=1\\i\neq j}^r a_ja_i$$
I try to use the formula $$\frac{a_i+a_j}{2}\geq \sqrt{a_ia_j}$$
however, I still cannot obtain a clear form. It also seems that we could combine some terms together; however, I cannot fully understand.
Want to show $\sum_{i=1}^r (r-1)a_i^2 \geq\sum_{i,j=1\\i\neq j}^r a_ja_i $
Since $\frac1{r}\sum_{i=1}^r a_i \le \left(\frac1{r}\sum_{i=1}^r a_i^2\right)^{1/2} $, $\left(\frac1{r}\sum_{i=1}^r a_i\right)^2 \le \frac1{r}\sum_{i=1}^r a_i^2 $, so that $\left(\sum_{i=1}^r a_i\right)^2 \le r\sum_{i=1}^r a_i^2 $.
Note: In response to a request for a proof of this, see this article: https://en.wikipedia.org/wiki/Generalized_mean
Therefore
$\begin{array}\\ \sum_{i,j=1\\i\neq j}^r a_ja_i &=\sum_{i=1}^r\sum_{j=1, j \ne i}^r a_ja_i\\ &=\sum_{i=1}^r\left(\sum_{j=1}^r a_ja_i-a_i^2\right)\\ &=\sum_{i=1}^r\sum_{j=1}^r a_ja_i-\sum_{i=1}^ra_i^2\\ &=\sum_{i=1}^ra_i\sum_{j=1}^r a_j-\sum_{i=1}^ra_i^2\\ &=\left(\sum_{i=1}^ra_i\right)^2-\sum_{i=1}^ra_i^2\\ &\le r\sum_{i=1}^r a_i^2-\sum_{i=1}^ra_i^2\\ &=(r-1)\sum_{i=1}^r a_i^2\\ \end{array} $