Let
![ABC](/media/m/a/c/7/ac75dca5ddb22ad70f492e2e0a153f95.png)
be a triangle for which there exists an interior point
![F](/media/m/3/e/8/3e8bad5df716d332365fca76f53c1743.png)
such that
![\angle AFB=\angle BFC=\angle CFA](/media/m/d/e/0/de09e506391e79ca5125b204ab80ca22.png)
. Let the lines
![BF](/media/m/4/d/4/4d4fbb8bdcff87c5df52488beb896501.png)
and
![CF](/media/m/6/7/0/670c216bc8a05762a60542376587c5fc.png)
meet the sides
![AC](/media/m/6/4/7/647ef3a5d68f07d59d84afe03a9dc655.png)
and
![AB](/media/m/5/2/9/5298bd9e7bc202ac21c423e51da3758e.png)
at
![D](/media/m/7/0/0/7006c4b57335ab717f8f20960577a9ef.png)
and
![E](/media/m/8/b/0/8b01e755d2253cb9a52f9e451d89ec11.png)
respectively. Prove that
%V0
Let $ABC$ be a triangle for which there exists an interior point $F$ such that $\angle AFB=\angle BFC=\angle CFA$. Let the lines $BF$ and $CF$ meet the sides $AC$ and $AB$ at $D$ and $E$ respectively. Prove that
$$AB+AC\geq4DE.$$