Let $a, b, c$ be the lengths of some triangle's sides, $p, r$ be the semiperimeter and the inradius of triangle. Prove an inequality $\sqrt{\frac{ab(p- c)}{p}} +\sqrt{\frac{ca(p- b)}{p}} +\sqrt{\frac{bc(p-a)}{p}} \ge 6r$ (D.Shvetsov)
Problem
Source: 2009 Sharygin Geometry Olympiad Final Round problem 1 grade 10
Tags: geometric inequality, geometry
28.07.2018 09:39
Hello. We know that $pr=2E$ so we have $r=\dfrac{p}{2E}$ and by substituting that we obtain that it is sufficient to prove $\sum \sqrt{ab(p-c)p} \geqslant 3E$. Now, we know from Heron's Formula that $E=\dfrac{\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}}{4}$. We make the Ravi substition, i.e. $a+b=x, b+c=y, c+a=z$. After the maniputations, we have to prove that $\sum \sqrt{\dfrac{(x+y)(y+z)}{xz}} \geqslant 6$. Now, $ \sum \sqrt{\dfrac{(x+y)(y+z)}{xz}} =\sum \sqrt{(1+\dfrac{y}{x})(1+\dfrac{y}{z})} \geqslant \sum (1+\dfrac{y^2}{xz})$, so it suffices to prove $\sum \dfrac{x^2}{yz} \geqslant 3$, which follows from the AM-GM inequality.
26.11.2024 16:26
Orestis_Lignos wrote: Hello. We know that $pr=2E$ so we have $r=\dfrac{p}{2E}$ and by substituting that we obtain that it is sufficient to prove $\sum \sqrt{ab(p-c)p} \geqslant 3E$. Now, we know from Heron's Formula that $E=\dfrac{\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}}{4}$. We make the Ravi substition, i.e. $a+b=x, b+c=y, c+a=z$. After the maniputations, we have to prove that $\sum \sqrt{\dfrac{(x+y)(y+z)}{xz}} \geqslant 6$. Now, $ \sum \sqrt{\dfrac{(x+y)(y+z)}{xz}} =\sum \sqrt{(1+\dfrac{y}{x})(1+\dfrac{y}{z})} \geqslant \sum (1+\dfrac{y^2}{xz})$, so it suffices to prove $\sum \dfrac{x^2}{yz} \geqslant 3$, which follows from the AM-GM inequality. P is not perimeter its semiperimeter , so pr=E
08.12.2024 10:42
From $\cos{\frac{A}{2}} = \sqrt{\frac{p(p-a)}{bc}}$ , we get $\sqrt{\frac{bc}{p}} = \frac{\sqrt{p-a}}{\cos{\frac{A}{2}}}$ . Therefore,$\sqrt{\frac{bc(p-a)}{p}} = \frac{(p-a)}{\cos{\frac{A}{2}}}$ and, ${\frac{1}{r}}{\sum{\sqrt{\frac{bc(p-a)}{p}}}}$ =$\sum{\frac{{(p-a)}}{r\cos{\frac{A}{2}}}}$ =$\sum{\frac{\cot{\frac{A}{2}}}{\cos{\frac{A}{2}}}}$ =$\sum{\frac{1}{\sin{\frac{A}{2}}}}$ and the stated inequality will follow if we can show that this last sum is greater than or equal to 6. That this last sum is greater or equal to 6 follows from $\sum{\frac{1}{\sin{\frac{A}{2}}}} \geq$ $\frac{9}{\sum{\sin{\frac{A}{2}}}}$, and Jensen's inequality $\sum{\sin{\frac{A}{2}}}$ $\leq$ $\frac{3}{2}$.
13.12.2024 12:10
Another proof that $\sum{\frac{1}{\sin{\frac{A}{2}}}}\geq6$ From $ \sin{\frac{A}{2}} = \sqrt{\frac{(p-b)(p-c)}{bc}}$ and $ (p-b) +(p-c) \geq 2\sqrt{(p-b)(p-c)}$ , we get $\sum{\frac{1}{\sin{\frac{A}{2}}}} = \sqrt{\frac{bc}{(p-b)(p-c)}}$ $\geq 2\sum{\frac{\sqrt{bc}}{(p-b) +(p-c)}}$ =$2\sum {\frac{\sqrt{bc}}{a}}\geq 2.3 =6$ ,where the last inequality is AM-GM and $(p-b) +(p-c) = 2p -b-c =a.$