Question

If a + b + c = 0, then prove that $\frac{(b^2+c^2)}{3bc}+\frac{(c+a{)}^2}{3ac}+\frac{(a+b{)}^2}{3ab}=\frac{1}{3}$

Answer 1

$a+b+c=0,$ then prove that $\frac{(b^2+c^2)}{3bc}+\frac{(c+a^2)}{3ac}+\frac{(a+b{)}^2}{3ab}=1/3$

$\frac{(b^2+c^2)}{3bc}+\frac{(c+a{)}^2}{3ac}+\frac{(a+b{)}^2}{3ab}=\frac{(b^2+c^2)}{3bc}+\frac{b^2}{3ac}+\frac{c^2}{3ab}$ since $a+b+c=0$

$=\frac{a(b^2+(c^2)+b(b^2)+c(c^2)}{3abc}$

$=\frac{a{b}^2+a{c}^2+b^3+c^3}{3abc}$

$=\frac{b^2(a+b)+c^2(a+b)}{3abc}$

$=\frac{b^2(-c)+c^2(-b)}{3abc}$

$=\frac{-bc(b+c)}{3abc}$

$=\frac{-bc(-a)}{3abc}$

$=\frac{abc}{3abc}$

$=\frac{1}{3}$