# 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.

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In the extreme case where $a+b \approx c$, the above term is close to zero. – Patrick Li Nov 18 '12 at 23:16
Just curious where this problem came from. Is it related to Heron's formula in some way? – Mike Nov 18 '12 at 23:35
I dont know where it comes from,I was asked by my brother – p.s Nov 18 '12 at 23:39
@Mike Yes, it is. And very related to the inradius ;) – Sawarnik May 6 '14 at 6:56

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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can you show it in a classical way, I need it for high school 1st year students? – p.s Nov 18 '12 at 23:21
OK, I'll do that. – Yury Nov 18 '12 at 23:23
what if a+b+c is fixed some p. – p.s Nov 18 '12 at 23:31
again it would be when a=b=c,but this way wouldnt work – p.s Nov 18 '12 at 23:32
just multiply all parameters $a$, $b$, $c$ and $\varepsilon$ by $p$ – Yury Nov 18 '12 at 23:33

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).

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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.

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