If `a_1,a_2,a_3,...,a_n in RR ` and all are non-negative, how can I prove this?

`(sqrt(a_1a_2)+sqrt(a_1a_3)+....+sqrt(a_1a_n))+`

`(sqrt(a_2a_3)+...+sqrt(a_2a_n))+`

`...`

`+(sqrt(a_(n-2)a_(n-1))+sqrt(a_(n-2)a_n))+`

`sqrt(a_(n-1)a_n)<=(n-1)/(2)(a_1+a_2+...+a_n)`

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I tried to edit your original attempt at writing the inequality by splitting it up over several lines. It's still a little awkward looking, but I think it's still pretty readable. I also made the assumtion that all the numbers are non-negative, since that turns out to be necessary. For the proof, we need the AM-GM inequality, which states that for any non-negative `a_j,a_k` we have `sqrt(a_ja_k)<=(a_j+a_k)/2.` Using this for every square root that occurs, we see that the left hand side of the inequality is less than or equal to

`((a_1+a_2)/2+(a_1+a_3)/2+...+(a_1+a_n)/2)+`

`((a_2+a_3)/2+(a_2+a_4)/2+...+(a_2+a_n)/2)+`

`...+`

`((a_(n-2)+a_(n-1))/2+(a_(n-2)+a_n)/2)+`

`(a_(n-1)+a_n)/2.`

Here there are `n-1` occurrences of `a_1/2,` all in the first line. There are also `n-1` occurrences of `a_2/2,` one in the first line and `n-2` in the second line. This pattern continues, and there are `n-1` occurrences of each number `a_j/2.` Thus the intimidating sum above simplifies to `(n-1)((a_1)/2+(a_2)/2+...+(a_n)/2),` which is what we needed to show.

**Sources:**

It didn't let me write the entire exercise:

sqrt(a1a2)+sqrt(a1a3)+....+sqrt(a1an)+sqrt(a2a3)+....+sqrt(an-1an)<=(n-1)/(2)(a1+a2+...+an)

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