<s>[INST] Area of a triangle whose each side is less than 2 and greater than1. What is the area of a triangle if each of its sides is greater than 1 and less than 2?
My Try:Let a,b,c be the sides of triangle,then
Area$=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{\frac{a+b+c}{2}\frac{b+c-a}{2}\frac{a+c-b}{2}\frac{a+b-c}{2}}=\frac{1}{4}\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}$
$1< a,b,c<2$
So  $3< a+b+c<6$ and $0< a+b-c,a+c-b,b+c-a<3$
Therefore  $0< Area<\frac{9}{4}\sqrt{2}$
Am I correct? Please also tell me some other method to solve this.
 [/INST] You can get arbitrarily small area, since you can get arbitrarily close to the degenerate triangle $a=b=1$, $c=2$, which has zero area.

The maximal possible area is obtained when forming equilateral triangle with $a=b=c=2$. (I.e., the area is $\sqrt3$.)
One of possible arguments is the following:
If two sides are shorter than $2$ units, you can increase the area by leaving the third side the same as it was and changing the height of the triangle by some small $\varepsilon>0$. (Which also increases the length of the two sides, but if the change is small enough, you still can keep those lengths $\le2$.) 
So it suffices to look at isosceles triangles with $b=c=2$ and $1\le a \le 2$. This means that we want to maximize the expression $\frac{a\sqrt{4-\frac{a^2}4}}2=\frac{\sqrt{a^2\left(4-\frac{a^2}4\right)}}2$. Since the function $t\mapsto t\left(4-\frac{t}4\right)$ is increasing for $t\in[1,4]$, the maximum is attained for $a=2$.
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