Prove this inequality in complex domain (5) Let $z_{1},z_{2},z_{3}\in C$ 
show that
$$|z_{1}+z_{2}+z_{3}|^2+|(z_{1}-z_{2})(z_{1}-z_{3})|+|(z_{2}-z_{3})(z_{2}-z_{1})|
+|(z_{3}-z_{1})(z_{3}-z_{2})|\le 3(|z_{1}|^2+|z_{2}|^2+|z_{3}|^2)$$
Iif prove following inequality,Use Cauchy-Schwarz inequality we have
$$(|z_{1}+z_{2}+z_{3}|^2\le (|z_{1}|+|z_{2}|+|z_{3}|)^2\le 3(|z_{1}|^2+|z_{2}|^2+|z_{3}|^2)$$
 A: In the following, I will instead of a complex number $z=x+ I y$ use the vector ${\bf v}\equiv (x,y)$ ($x,y$ being the $Ox$ and $Oy$ projections of vector $\vec v$) for allowed operations: eg, the modulus of a complex number is the same as the magnitude of the equivalent vector. Note that $v^2$ is now the magnitude squared of the vector $\bf{v}$.
Firstly, let us define the center-of-mass (COM) vector ${\bf d}=\frac{{\bf v}_1+{\bf v}_2+{\bf v}_3}{3}$ and ${\bf d}_i = {\bf v}_i -{\bf d}$ the relative positions to the  COM of each of the points represented by ${\bf v}_i$ ($i=1,2,3$). We can now write the inequality as: 
$$9 d^2 + d_1d_2+d_2d_3+d_3d_1 \le 3\sum_{i=1,2,3}\vert {\bf d}+{\bf d}_i\vert^2,$$
which by using $\vert {\bf a}+{\bf b} \vert^2=a^2+b^2+2{\bf a \cdot b}$ (the last term being the scalar product of the vectors), one gets:
$$9 d^2 + d_1d_2+d_2d_3+d_3d_1 \le 3(3d^2+2{\bf d\cdot}\sum_{i=1,2,3}{\bf d}_i,+\sum_{i=1,2,3}d_i^2)$$. Using $\sum_{i=1,2,3}\, {\bf d}_i=0$, we have left to prove only that $3(d_1^2+d_2^2+d_3^2)\ge d_1d_2+d_2d_3 + d_3d_1$, which is equivalent to $\sum_{i,j,i\ne j} (d_i - d_j)^2/2 \ge 0$ (all indexes being $1,2,3$). 
A: Let $O$, $A$, $B$ and $C$ be points in the plain and $M$ is a gravity canter of $\Delta ABC$.
Hence, 
$3(OA^2+OB^2+OC^2)=9OM^2+AB^2+AC^2+BC^2\geq$
$\geq9OM^2+AB\cdot AC+AB\cdot BC+AC\cdot BC$ and we are done!
