If `triangle` is the area and 2s is the perimeter of `triangleABC`, then prove that `triangle le s^(2)/(3sqrt(3))`

Explanation

2s=a+b+c, `triangle^(2) = s(s-a)(s-b)(s-c)`
`therefore` In `triangleABC`,
`a + b gt c rArr a+b-c gt 0`
`rArr 2s-2c gt 0`
`rArr s-c gt 0`
Similarly, `s-b gt 0` nad `s-a gt 0`
`therefore` Using A.M. `ge` G.M.
`rArr ((s-a)+(s-b)+(s-c))/(3) ge ((s-a)(s-b)(s-c))^(1//3)`
`rArr (3s-(a+b+c))/(3) ge ((s-a)(s-b)(s-c))^(1//3)`
`rArr s/3 ge (Delta^(2)/s)^(1//3) rArr s^(3)/27 ge Delta^(2)/s`
`rArr s^(4) ge 27Delta^(2)`
`rArr s^(2)/(3sqrt(3)) ge Delta`
`Delta le s^(2)/(3sqrt(3))`