Problem:
Of all triangles with a given perimeter, find the one with maximum area.
Solution:
Consider an arbitrary triangle with side lengths a, b, c and perimeter 
Now, the arithmetic mean geometric mean inequality gives
![\sqrt[3]{s(s-a)(s-b)(s-c)} \leq \frac{(s-a)+(s-b)+(s-c)}{3}=\frac{s}{3}](https://s0.wp.com/latex.php?latex=%5Csqrt%5B3%5D%7Bs%28s-a%29%28s-b%29%28s-c%29%7D+%5Cleq+%5Cfrac%7B%28s-a%29%2B%28s-b%29%2B%28s-c%29%7D%7B3%7D%3D%5Cfrac%7Bs%7D%7B3%7D&bg=ffffff&fg=000&s=0 "\sqrt[3]{s(s-a)(s-b)(s-c)} \leq \frac{(s-a)+(s-b)+(s-c)}{3}=\frac{s}{3}")
Therefore,
where inequality holds if and only if 
Thus, the area of any triangle with perimeter 2s does not exceed 
More later,
Nalin Pithwa
PS: Ref: Geometric Problems on Maxima and Minima by Titu Andreescu et al.
