If $2(bc^2+ca^2+ab^2) = b^2c+c^2a+a^2b+3abc, $ then prove that triangle $ABC$ is equilateral In a triangle $ABC,$ if $2(bc^2+ca^2+ab^2) = b^2c+c^2a+a^2b+3abc, $ then prove that triangle 
$ABC$ is equilateral 
$\displaystyle \frac{2(bc^2+ca^2+ab^2)}{abc} = \frac{b^2c+c^2a+a^2b+3abc}{abc}$
$\displaystyle 2\left(\frac{c}{a}+\frac{a}{b}+\frac{b}{c}\right) = \left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)+3$
let $\displaystyle \frac{a}{b} = x,\frac{b}{c}=y,\frac{c}{a}=z$
so $\displaystyle 2(x+y+z) = \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3$
wan,t be able to proceed after that, could some help me

Because 
$$0=\sum\limits_{cyc}(2b^2a-a^2b-abc)=\sum\limits_{cyc}(a^3-abc-(b^3-2b^2a+a^2b))=$$
$$=(a+b+c)\sum\limits_{cyc}(a^2-ab)-\sum\limits_{cyc}b(a-b)^2=$$
$$=\sum\limits_{cyc}(a-b)^2\left(\frac{1}{2}(a+b+c)-b\right)=\frac{1}{2}\sum\limits_{cyc}(a-b)^2(a+c-b).$$
Id est, the given it's
$$(a-b)^2(a+c-b)+(b-c)^2(b+a-c)+(c-a)^2(c+b-a)=0,$$
which says that $a=b=c$.
Done!
