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
 A: Since $a,b,c$ are sides of a triangle we may write $(a,b,c) = (y+z,z+x,x+y)$
with $x,y,z>0$.  The difference between the two sides is then
$$
x^3 + y^3 + z^3 \; + \; x^2 y + y^2 z + z^2 x
- 2 (xy^2 + yz^2 + zx^2).
$$
By cyclic symmetry we may assume $x \geq z$ and $y \geq z$, and write
$x = (1 + \alpha) z$ and $y = (1 + \beta) z$ with $\alpha,\beta \geq 0$
to find $z^3$ times
$$
2(\alpha^2 - \alpha\beta + \beta^2) + \beta(\alpha-\beta)^2 + \alpha^3,
$$
in which each term is nonnegative, and all are zero iff
$\alpha = \beta = 0$ $-$ which in turn is equivalent to $x=y=z$,
and thus $a=b=c$, so triangle $ABC$ is equilateral as claimed.
$\Box$
[Added later: The solution above applies a general technique for such 
problems; but I see that for the present question 
Hari Shankar posted (in a comment) a link to an
AoPS item
that gives an even simpler conclusion that retains the cyclic symmetry:
$$
x^3 + y^3 + z^3 \; + \; x^2 y + y^2 z + z^2 x
- 2 (xy^2 + yz^2 + zx^2) = x(x-z)^2 + y(y-x)^2 + z (z-y)^2,
$$
again with all terms nonnegative; so equality implies that
they all vanish and the rest follows as before.  That link also
gives the example $(a:b:c) = (10:3:24)$ of a rational point on the cubic
$2(bc^2 + ca^2 + ab^2) = b^2c + c^2a + a^2b + 3abc$ with positive variables
that do not satisfy the triangle inequality.]
A: 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!
A: $x+y+z\geq 3(xyz)^{\frac{1}{3}}$,when $x=y=z$ ,equal sign is established.So:$$2\sum\frac{c}{a}\geq\sum\frac{c}{a}+3$$ i.e.$$\frac{c}{a}+\frac{b}{c}+\frac{a}{b}\leq\frac{a}{c}+\frac{b}{a}+\frac{c}{b}$$Without loss of generality, we assume:$c\geq b\geq a$,so $\frac1{a}\geq\frac1{b}\geq\frac1{c},$then we have:$$\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\leq\frac{c}{a}+\frac{b}{c}+\frac{a}{b},$$so $a=b=c$.I'm using an incomplete sort to derive a contradiction.
