Geometrical Inequalities I couldnt solve the following: we need to minimize $$\sqrt{\frac{(a+b-c)(b+c-a)(a+c-b)}{(a+b+c)}},$$ where a,b,c are sides of a triangle.
 A: This expression can be arbitrarily close to $0$. Let $a=2\varepsilon$, $b = c = 1/2 - \varepsilon$. There is a triangle with sides $a$, $b$, and $c$. As $\varepsilon$ goes to $0$, the expression approaches $0$. Specifically, we have
$$\sqrt{\frac{(a+b-c)(b+c-a)(a+c-b)}{(a+b+c)}} 
= \sqrt{\frac{2\varepsilon \cdot (1 - 4\varepsilon)\cdot 2\varepsilon}{1}} \leq \sqrt{2\varepsilon \cdot 2\varepsilon} = 2\varepsilon.$$
This expression can be arbitrarily close to $0$.
The expression is maximized when $a=b=c$.
A: Also, by Heron's formula the expression in question is Area/4 Perimeter. So of course it can be arbitrary close to zero. (Note that the maximal area given fixed perimeter is for the equilateral triangle, as can be shown geometrically rather easily).  
A: Yes this expression can get arbitrarily close to zero.
By multiplying both the numerator and the denominator of the fraction by (a+b+c) and simplifying and plugging in A=rs (r is in-radius s is semi-perimeter) we have our eexpression is equivalent to
$$ 2r$$
Therefore it can get arbitrarily small.
