Show that point $P $ and $\Delta ABC$ satisfy the given condition 
Let ABC be a triangle and $P$ be any point in complex plane. Then show that 
  $$1.BC^2+CA^2+AB^2 \le 3 (PA^2+PB^2+PC^2)$$
  $$2.BC^2+CA^2+AB^2 = 3 (PA^2+PB^2+PC^2) \iff \text {P is centroid of}  \space\Delta ABC$$

I tried to solve it by assuming point $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3),P (x,y)$
$$2({x_1}^2+{x_2}^2+{x_3}^2)+2({y_1}^2+{
y_2}^2+{y_3}^2)-2({x_1}{x_2}+{x_2}{x_3}+{x_1}{x_3})-2({y_1}{y_2}+{y_2}{y_3}+{y_1}{y_3})=3({x_1}^2+{x_2}^2+{x_3}^2+{y_1}^2+{
y_2}^2+{y_3}^2+3x^2+3y^2-2x({x_1}+{x_2}+{x_3})-2y({y_1}+{y_2}+{y_3}))$$
I am not able to conclude anything from this. 

Your expression is wrong. At least, you must point to what is the LHS and RHS. Let me guess. You mean LHS is $AB^2 + BC^2 + CA^2$ and RHS is $3(PA^2 + PB^2 + PC^2)$, but we have $$LHS = AB^2 + BC^2 + CA^2 = 2\sum x_i^2 + 2\sum y_i^2 -\sum 2x_ix_j - 2\sum y_iy_j,$$
$$RHS = 3(PA^2 + PB^2 + PC^2) = 3[3(x^2+y^2) + \sum x_i^2  + \sum y_i^2 - 2x\sum x_i - 2y \sum y_i]$$
Then, you can see easily that $LHS < RHS$. For example, 
$$RHS - LHS = 9(x^2+y^2) + \sum x_i^2 + \sum y_i^2 + 2\sum x_ix_j + 2\sum y_iy_j - 2x\sum x_i -2y\sum y_i$$
$$ = 9(x^2+y^2) + (\sum x_i)^2 + (\sum y_i)^2 - 2 x\sum x_i - 2y\sum y_i$$
$$ = 8(x^2+y^2) + (x - \sum x_i)^2 + (y-\sum y_i)^2$$
