triangular inequality [sides and area]
Posted: Sat May 02, 2015 1:15 am
Prove that for any triangle $ABC$ with sides $a,b,c$ and area $A$,
$a^{2}+b^{2}+c^{2}\geq 4\sqrt{3}A$
$a^{2}+b^{2}+c^{2}\geq 4\sqrt{3}A$
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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.For any $\triangle ABC$ with side lengths $a,b,c$ and area $\Delta$ we have $\displaystyle\sum_{\text{cyc}} a^2-\sum_{\text{cyc}}(b-c)^2\ge 4\sqrt 3\Delta$.