Geometric inequality : In a equilateral , $ 4(PD^2 +PE^2 + PF^2) \ge PA^2 + PB^2 + PC^2 $ In a equilateral $ABC$ , 
Prove
$$ 4(PD^2 + PE^2 + PF^2 ) \ge PA^2 + PB^2 + PC^2 $$
$(PD, PE, PF   $ are distance to sides, and P is located same plane with $ABC$)
I know that;
$$PA + PB \ge PC , $$
$$ PB + PC  \ge  PA , $$
$$ PC + PA  \ge PB $$
and When P is located on the circumcircle , $'D, E, F$  is colinear.'  But these are not useful to solve the problem. I want some hints. Thank you.
 A: Let $P(x,y),$ $A(1,0)$, $B\left(-\frac{1}{2},\frac{\sqrt3}{2}\right)$ and $C\left(-\frac{1}{2},-\frac{\sqrt3}{2}\right)$.
Thus, $$d(P, AB)=\frac{|x+\sqrt3y-1|}{2}$$
$$d(P,AC)=\frac{|x-\sqrt3y-1|}{2},$$
$$d(P,BC)=\left|x+\frac{1}{2}\right|$$ and we need to prove that
$$(2x+1)^2+(x+\sqrt3y-1)^2+(x-\sqrt3y-1)^2\geq$$
$$\geq(x-1)^2+y^2+\left(x+\frac{1}{2}\right)^2+\left(y-\frac{\sqrt3}{2}\right)^2+\left(x+\frac{1}{2}\right)^2+\left(y+\frac{\sqrt3}{2}\right)^2,$$
which is $$x^2+y^2\geq0.$$
A: Following is an approach using barycentric coordinates.
WOLOG, we will assume the equilateral triangle $ABC$ has unit side length.
We will also assume $D, E, F$ lie on sides $BC, CA, AB$ respectively.
If $(x,y,z)$ is the barycentric coordinates of $P$. i.e.
$$\vec{P} = x\vec{A} + y\vec{B} + z\vec{C}\quad\text{ and }\quad x + y + z = 1$$
then on LHS, we have 
$PD = \frac{\sqrt{3}}{2} |x|$,
$PE = \frac{\sqrt{3}}{2} |y|$ and $PF = \frac{\sqrt{3}}{2} |z|$. This leads to
$${\rm LHS} \stackrel{def}{=} 4(PD^2 + PE^2 + PF^2) = 3(x^2+y^2+z^2)$$
On the RHS, we will use the fact when $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are the barycentric coordinates of two points $Q,R$ with respect to
a triangle of sides $a, b, c$, their distance is given by following formula:
$$QR^2 = -a^2(y_1-y_2)(z_1-z_2) - b^2(z_1-z_2)(x_1-x_2) -c^2(x_1-x_2)(z_1-z_2)$$
Apply this to segments $PA$, $PB$ and $PC$, we get
$$\begin{align}
PA^2 &= y+z - (xy+yz+zx)\\
PB^2 &= x+z - (xy+yz+zx)\\
PC^2 &= x+y - (xy+yz+zx)
\end{align}
$$
This leads to
$$\begin{align}
{\rm RHS} &\stackrel{def}{=} PA^2 + PB^2 + PC^2\\
&= 2(x+y+z) - 3(xy+yz+zx) = 2 - 3(xy+yz+xz)\\
&= 2(x+y+z)^2-3(xy+yz+zx)
= 2(x^2+y^2+z^2) + (xy+yz+zx)
\end{align}
$$
Combine these two expressions, we find
$${\rm LHS} - {\rm RHS}
= (x^2+y^2+z^2) - (xy+yz+zx)
= \frac12\left((x-y)^2 + (y-z)^2 + (z-x)^2\right) \ge 0$$
This establish the inequality:
$$4(PD^2+PE^2+PF^2) = {\rm LHS} \ge {\rm RHS} = PA^2+PB^2+PC^2$$
and the inequality is strict unless $x = y = z$, i.e. when $P$ is not the centroid.
A: If $ABC$ is an equilateral triangle, $P$ a point inside it and $D,E,F$ its projections on the sides, we may assume $AB=1$ without loss of generality and notice that $PD+PE+PF=\frac{1}{2}\sqrt{3}$ by Viviani's theorem and $PA^2+PB^2+PC^2 = 3PG^2+1$ by the parallel axis theorem, where $G$ is the center of $ABC$. In general, given a triangle $UVW$, the point $Q$ minimizing $QU^2+QV^2+QW^2$ is the centroid of $UVW$. So, if we denote as $Q$ the centroid of the pedal triangle of $DEF$ we have
$$ PD^2+PE^2+PF^2 \geq QD^2+QE^2+QF^2. $$
On the other hand, by the intercept theorem $Q$ is exactly the midpoint of $PG$, hence
$$ QD^2+QE^2+QF^2 = QA'^2+QB'^2+QC'^2 $$
where $A'B'C'$ is the medial triangle of $ABC$. The claim is now trivial, since $ABC$ and $A'B'C'$ are similar and share the same centroid.

