Question

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

Answer 1

Given

$a+b+c=0$

$a+b=-c$

Cubing on both sides

$(a+b{)}^3=(-c{)}^3$

$a^3+b^3+3\left(ab\right)(a+b)=-c^3$

$a^3+b^3+3ab(-c)=c^3$

$a^3+b^3-3abc=-c^3$

$a^3+b^3+c^3=3abc$

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

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

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

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

$=\frac{3abc}{3abc}=1$