Let $a_1,\,a_2,\,a_3,\,\dots,\,a_n$ be $n$ arbitrary real numbers. Then, how to prove the following inequality? $$\left({\sum_{k=1}^n{a_k}}\right)^2 \leq \left(n-1\right)\left(\sum_{k=1}^n{a_k^2}+2a_ia_j\right)$$
Here, $i,\,j\in\left\{1,\,2,\,\dots,\,n\right\}$ are arbitrary natural numbers.
Without loss of generality we may assume that $i=1$ and $j=2$. Then we have $$(n-1)\left(\sum_{k=1}^n{a_k}^2+2a_1a_2\right)=(n-1)({a_3}^2+...+{a_n}^2+(a_1+a_2)^2).\ \ (1)$$
From the Cauchy-Schwarz inequality for the $(n-1)$-tuples $(1,...,1)$ and $(a_3,...,a_n,a_1+a_2)$ we obtain $$(n-1)({a_3}^2+...+{a_n}^2+(a_1+a_2)^2)\geq \left(\sum_{k=3}^na_k+(a_1+a_2)\right)^2=\left(\sum_{k=1}^na_k\right)^2.\ \ (2)$$
Combine $(1)$ and $(2)$ and you are done.