Question

If a+b+c=0, then prove that (b+c)²/6bc + (c+a)²/6ac + (a+b)²/6ab = 1/2

Answer 1

$\large\text{ \boxed{\bold{[Topic]}} }$

Polynomial identities

Identity of three variables

$\large\cdots\longrightarrow\boxed{a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)}$

This equals zero if and only if $\largea=b=c\text{ or }a+b+c=0.$

Given condition: -

$\large\text{ \bullet\ a+b+c=0\Longrightarrow\boxed{a^{3}+b^{3}+c^{3}=3abc} }$

To prove: -

$\large\text{ \bullet\ \dfrac{(b+c)^{2}}{6bc}+\dfrac{(c+a)^{2}}{6ca}+\dfrac{(a+b)^{2}}{6ab}=\dfrac{1}{2} }$

$\large\text{ \boxed{\bold{[Step\ 1.]}} }$

We know that -

$\large\cdots\longrightarrow\boxed{a+b+c=0.}$

Consequently, -

$\large\cdots\longrightarrow b+c=-a,\ c+a=-b,\ a+b=-c.$

$\large\text{ \boxed{\bold{[Step\ 2.]}} }$

The equation to prove is -

$\large\text{ \cdots\longrightarrow\dfrac{(b+c)^{2}}{6bc}+\dfrac{(c+a)^{2}}{6bc}+\dfrac{(a+b)^{2}}{6bc}=\dfrac{1}{2} }$

LHS

$\large\text{ \cdots\longrightarrow\boxed{\begin{aligned}&\dfrac{(b+c)^{2}}{6bc}+\dfrac{(c+a)^{2}}{6ca}+\dfrac{(a+b)^{2}}{6ab}\\\\&=\dfrac{(-a)^{2}}{6bc}+\dfrac{(-b)^{2}}{6ca}+\dfrac{(-c)^{2}}{6ab}\\\\&=\dfrac{a^{3}+b^{3}+c^{3}}{6abc}\\\\&=\dfrac{3abc}{6abc}\\\\&=\dfrac{1}{2}\end{aligned}} }$

Now, the LHS and RHS are equal.

$\large\text{ \boxed{\bold{[Final\ answer]}} }$

We finish our answer.

$\large\text{ \cdots\longrightarrow\boxed{\bold{\dfrac{(b+c)^{2}}{6bc}+\dfrac{(c+a)^{2}}{6ca}+\dfrac{(a+b)^{2}}{6ab}=\dfrac{1}{2}.}} }$