I am currently trying to figure out if it's possible to give a better lower bound to the area of a triangle in terms of its semi-perimeter (or perimeter).
If the sides of a triangle are $a$, $b$ and $c$, and the semiperimeter is $s$, then using Heron's formula, I get: $16T^2 = s(s - a)(s - b)(s - c)$
Since all of $(s-a)$, $(s-b)$, $(s-c)$ are smaller than $s$, by substituting $s$ I can give an upper bound of $16T^2 < s^4$, and from there $T^2 < s^4/16$ pretty easily. On the other hand, to get a lower bound, other than simply eliminating the terms I mentioned above to get $s < 16T^2$ and then $T^2 > s/16$, I don't see what else I can do...
Surely, this lower bound can be improved right?
EDIT: I am also assuming the triangles I am working with, all have integer sides and an angle of 60 degrees.
The greatest lower bound of $T$, for a given $s$, is $0.$
Given $s>0,$ take $t \in (0,\pi /2).$ Let $U=s/(1+\cos t)$
Take isosceles triangle $ABC$ with $AB=AC=U$ and $\angle ABC=\angle ACB=t.$ Then $BC=2(AB)\cos \angle ABC=2U\cos t .$
So the semi-perimeter of $ABC$ is $U(1+\cos t)=s.$
The area of $ABC$ is $$T= \frac {1}{2}(BC)(AB\sin \angle ABC)=\frac {1}{2} (2U\cos t)(U\sin t)= $$ $$= U^2\cos t \sin t<U^2 \sin t<s^2\sin t$$ which can be arbitrarily small if $t$ is small enough