Question

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

Answer 1

Let us solve the LHS of the given expression $\frac{{(b+c)}^2}{3bc}+\frac{{(c+a)}^2}{3ac}+\frac{{(a+b)}^2}{3ab}$ as shown below:

$\frac{{(b+c)}^2}{3bc}+\frac{{(c+a)}^2}{3ac}+\frac{{(a+b)}^2}{3ab}=\frac{{a(b+c)}^2}{3abc}+\frac{{b(c+a)}^2}{3abc}+\frac{{c(a+b)}^2}{3abc}\left(TakingLCM\right)=\frac{a(b^2+c^2+2bc)}{3abc}+\frac{b(c^2+a^2+2ac)}{3abc}+\frac{c(a^2+b^2+2ab)}{3abc}(\because (x+y{)}^2=x^2+y^2+2xy)=\frac{a{b}^2+a{c}^2+2abc}{3abc}+\frac{b{c}^2+b{a}^2+2abc}{3abc}+\frac{c{a}^2+c{b}^2+2abc}{3abc}$

$=\frac{a{b}^2+a{c}^2+b{c}^2+b{a}^2+c{a}^2+c{b}^2+6abc}{3abc}=\frac{(b{a}^2+a{b}^2)+(c{b}^2+b{c}^2)+(c{a}^2+a{c}^2)+6abc}{3abc}=\frac{ab(a+b)+bc(b+c)+ac(a+c)+6abc}{3abc}=\frac{ab(-c)+bc(-a)+ac(-b)+6abc}{3abc}(Givena+b+c=0)$

$=\frac{-abc-abc-abc+6abc}{3abc}=\frac{-3abc+6abc}{3abc}=\frac{3abc}{3abc}=1$

Hence, $\frac{{(b+c)}^2}{3bc}+\frac{{(c+a)}^2}{3ac}+\frac{{(a+b)}^2}{3ab}=1$.