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.
Geometrical Inequalities
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0@Mike Yes, it is. And very related to the inradius ;) – 2014-05-06
3 Answers
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 $a=b=c$.
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0just multiply all parameters $a$, b, $c$ and $\varepsilon$ by $p$ – 2012-11-18
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).
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.