TITLE: Prove or disprove inequality: $2a^2 + 2b^2 + 3c^2 \ge 16P$
QUESTION [6 upvotes]: Let $a,b,c$ be the lengths of the sides of a triangle with area $P$ 
prove or disprove inequality:  $2a^2 + 2b^2 + 3c^2 \ge 16P$

REPLY [2 votes]: Here is a short algebraic proof.  Since both sides are homogeneous of degree $2$, we can rescale to the case where $c=1$.  Then  by translating $A$ to the origin and by rotating so that $B$ lies at $(1,0))$ , we can reduce to the case where $A=(0,0)$, $B=(1,0)$ and $C$ is $(p,q)$ for some $p$ and $q$.  The inequality then becomes
$$2(p^2+q^2) + 2((p-1))^2 + q^2) + 3 \geq 16 q.$$
This is,  by simple algebra, equivalent to the valid inequality 
$ (p-\frac 12 )^2+(q-1)^2 \geq 0$.