If $2(bc^2+ca^2+ab^2) = b^2c+c^2a+a^2b+3abc, $ then prove that triangle $ABC$ is equilateral

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.]

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!