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.

• @N.S.JOHN 12 standard.Why this if off topic ? – Aakash Kumar Oct 9 '16 at 15:17

Hint: with the usual notation where $z \in \mathbb{C}$ corresponds to point $Z$ in the complex plane, expand the squares e.g. $AB^2 = |a-b|^2 = (a-b)(\bar a - \bar b)$ then collect and regroup. The inequality reduces to:

$$RHS - LHS = |3 p - (a+b+c)|^2 \ge 0$$

which always holds true, with equality iff $p=\frac{a+b+c}{3}$ $\iff$ $P$ is the centroid of $\triangle ABC$.

Sketch of the proof: let for brevity the sums $\sum$ mean the sums for cyclic permutations of $(a,b,c)$ so that for example $\sum b \bar c = b \bar c + c \bar a + a \bar b$ and $\sum |p|^2 = 3 |p|^2$. Then:

\begin{align} RHS - LHS & = 3 \sum |p-a|^2 - \sum |b-c|^2 \\ & = 3 \sum (p-a)(\bar p - \bar a) - \sum (b-c)(\bar b - \bar c) \\ & = 9 |p|^2 - 3 \sum (p \bar a+ \bar p a) + 3 \sum |a|^2 - 2 \sum |a|^2 + \sum (b \bar c + \bar b c) \\ & = 9 |p|^2 - 3 \sum (p \bar a+ \bar p a) + \sum (b \bar c + \bar b c) + \sum |a|^2 \\ & = (3p - \sum a)(3 \bar p - \sum \bar a) \\ & = \left|3p - \sum a\right|^2 \quad \ge \quad 0 \end{align}

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,

$$RHS - LHS = 9(x^2+y^2) + \sum x_i^2 + \sum y_i^2 + 2\sum x_ix_j + 2\sum y_iy_j - 2x\sum x_i -2y\sum y_i$$

$$= 9(x^2+y^2) + (\sum x_i)^2 + (\sum y_i)^2 - 2 x\sum x_i - 2y\sum y_i$$

$$= 8(x^2+y^2) + (x - \sum x_i)^2 + (y-\sum y_i)^2$$