triangular inequality [sides and area]

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Tahmid
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triangular inequality [sides and area]

Unread post by Tahmid » Sat May 02, 2015 1:15 am

Prove that for any triangle $ABC$ with sides $a,b,c$ and area $A$,
$a^{2}+b^{2}+c^{2}\geq 4\sqrt{3}A$

Nirjhor
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Re: triangular inequality [sides and area]

Unread post by Nirjhor » Sun May 03, 2015 1:23 am

Weitzenbock's inequality has like more than ten/eleven different proofs. When I first confronted it, I instead found a stronger one which, surprisingly, is easier to prove.
For any $\triangle ABC$ with side lengths $a,b,c$ and area $\Delta$ we have $\displaystyle\sum_{\text{cyc}} a^2-\sum_{\text{cyc}}(b-c)^2\ge 4\sqrt 3\Delta$.
Set $x=b+c-a, ~y=c+a-b,~ z=a+b-c$. Then the left side factorizes as $xy+yz+zx$. And by Heron's formula $4\Delta=\sqrt{xyz(x+y+z)}$. So we have to prove that \[xy+yz+zx\ge \sqrt{3xyz(x+y+z)}\Leftrightarrow (xy+yz+zx)^2\ge 3(xy\cdot yz+yz\cdot zx+zx\cdot xy)\] which is trivial. Equality at equilateral triangle.
Last edited by Nirjhor on Sun May 03, 2015 1:59 am, edited 1 time in total.
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- Because all the poles are in Eastern Europe.


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sowmitra
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Re: triangular inequality [sides and area]

Unread post by sowmitra » Sun May 03, 2015 1:43 am

This was also set as Problem No. 2 in the 1961 IMO. :geek:
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Some-Angle Related Problems;

Tahmid
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Re: triangular inequality [sides and area]

Unread post by Tahmid » Sun May 03, 2015 12:22 pm

my solution is same as nirjhor :)

yeah sowmitra vaia this is IMO 1961/2 :|

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