There is point $P$ in a triangle $ABC$. $Q,R,S$ are the symmetric of $P$ with respect to the sides $AB,BC,CA$ respectively. I have to prove that the area of $ABC$ is $\geq$ than the area of $QRS$.
Any ideas?
P.S. I noticed that if $P$ is circumcenter, the area of two triangles is the same.
The area of $QRS$ is just four times the area of the pedal triangle of $P$.
By Euler's theorem, the area of the pedal triangle just depends on the distance from the circumcenter $O$: $$ [QRS]=\left(1-\frac{OP^2}{R^2}\right)[ABC] $$ hence the claim is trivial.