Consider a triangle $ABC$. A directly similar triangle $A_1B_1C_1$ is inscribed in the triangle $ABC$ such that $A_1,\;B_1\;,C_1$ are the interior points of the sides $AC,\;AB\;and\;BC$ respectively. Prove thar $$\frac{Area(\Delta A_1B_1C_1)}{Area(\Delta ABC)}\geq\frac{1}{cosec^2A+cosec^2B+cosec^2C}$$
I don't even know how to get started. Any ideas on how to solve the problem. Thanks.
Because the question raised in the comment was not answered, I assume that the “interior” point of a side means the “midpoint” of the side. This is one way that a “directly similar (inscribed) triangle A1B1C1” can be generated from triangle ABC.
Since $\frac{Area(\Delta A_1B_1C_1)}{Area(\Delta ABC)} = \frac {1}{4}$, what remain to be shown is
$\frac {1}{4}\geq\frac{1}{cosec^2A+cosec^2B+cosec^2C}$
That is, $cosec^2A+cosec^2B+cosec^2C \geq 4$
Or $\frac {1}{\sin^2(A)} + \frac {1}{\sin^2(B)} + \frac {1}{\sin^2(C)} \geq 4$
The rest is solved by experts (not me) in here.
