Are identities used here?

By using $A.M-G.M$ we get

$\sum_{ω=2}^{ω}\sqrt{a_{1}a_{ω}}\leq(\frac{(n-1)a_{1}+\sum_{ω=2}^{ω}a_{ω})}{2}$

$\sum_{ω=3}^{ω}\sqrt{a_{2}a_{ω}}\leq(\frac{(n-2)a_{2}+\sum_{ω=3}^{ω}a_{ω})}{2}$ . . . $\sqrt{a_{ω-1}a_{ω}}\leq\frac{a_{ω-1}+a_{ω}}{2}$

Adding all above equations we get

$2\sum_{I<J}\sqrt{a_{I}a_{J}}\leq(n-1)( \sum_{ω=1}^{ω}a_{ω})$ Now putting values we get the answer as $2015$ .

If $a_1,a_2,\ldots,a_n$ are positive numbers whose sum is $N$ , then $2\sum_{1 \le i < j \le n} \sqrt{a_ia_j} \; = \; \left(\sum_{i=1}^n \sqrt{a_i}\right)^2 - \sum_{i=1}^n a_i \; = \; \left(\sum_{i=1}^n \sqrt{a_i}\right)^2 - N$ Using the Cauchy-Schwarz Inequality, $2\sum_{1 \le i < j \le n} \sqrt{a_ia_j} \; \le \; \left(\sum_{i=1}^n 1^2\right)\left(\sum_{i=1}^n a_i\right) - N \; = \; nN - N \; = \; (n-1)N \;,$ with equality when $a_1=a_2=\ldots=a_n = \tfrac{N}{n}$ . The maximum value is thus $(n-1)N$ . In this case, where $n=14,N=155$ , the maximum is $2015$ .

The numbers are all independent. Hence the sum is maximized when each term is maximized. From AM-GM, terms are maximum when all the numbers are equal. That is

$a_i=a \forall i$

Hence 14a=155 There are 91 terms in the sum.

Hence the answer is $\frac{2\times 91 \times 155}{14}=2015$