Prove that for any triangle $ABC$ with sides $a,b,c$ and area $A$,

$a^{2}+b^{2}+c^{2}\geq 4\sqrt{3}A$

## triangular inequality [sides and area]

### Re: triangular inequality [sides and area]

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.

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 $\displaystyle\sum_{\text{cyc}} a^2-\sum_{\text{cyc}}(b-c)^2\ge 4\sqrt 3\Delta$.

Last edited by Nirjhor on Sun May 03, 2015 1:59 am, edited 1 time in total.

**- What is the value of the contour integral around Western Europe?**

- Zero.

- Why?

- Because all the poles are in Eastern Europe.

- Zero.

- Why?

- Because all the poles are in Eastern Europe.

Revive the IMO marathon.

### Re: triangular inequality [sides and area]

This was also set as Problem No. 2 in the 1961 IMO.

### Re: triangular inequality [sides and area]

my solution is same as nirjhor

yeah sowmitra vaia this is IMO 1961/2

yeah sowmitra vaia this is IMO 1961/2