TITLE: $PA^2\sin A+PB^2\sin B+PC^2\sin C$ is minimum if P is the incenter.
QUESTION [8 upvotes]: Let $ABC$ be a triangle and $P$ is a point in the plane of the triangle $ABC$.If $a,b,c$ are the lengths of sides $BC,CA,AB$ opposite to angles $A,B,C$ respectively then prove that

$PA^2\sin A+PB^2\sin B+PC^2\sin C$ is minimum if P is the incenter.
$PA^2\tan A+PB^2\tan B+PC^2\tan C$ is minimum if P is the orthocenter.
$PA^2+PB^2+PC^2$ is minimum if P is the centroid.
$PA^2\sin 2A+PB^2\sin 2B+PC^2\sin 2C$ is minimum if P is the circumcenter.

How do i have to formulate this expression into a form that is differentiable wrt a single variable or is there some different hidden approach will be used here.Please guide me,how should i solve this question?

REPLY [3 votes]: For 1, let $X,Y,Z$ be the projections of $P$ onto the sides of triangle $ABC$. Note that by the extended law of sines, $YZ=AP\sin A$, etc. Therefore our expression
$$
AP^2\sin A+BP^2\sin B+CP^2\sin C=AP\cdot YZ+BP\cdot ZX+CP\cdot XY\ge 2[AYPZ]+2[BZPX]+2[CXPY]=2[ABC]
$$
with equality when $P$ is the incenter, as then $AP\perp YZ$, etc.