i actually thought of a different method
2(bc^2+ca^2+ab^2)=b^2c+c^2a+a^2b+3abc
2abc(b/c+c/a+a/b)=abc(b/a+a/c+c/b+3)
put c/b=p
2(1/p+pb/a+a/b)=(b/a+a/pb+p+3)
put b/a=m
2(1/p+pm+1/m-1)=(m+1/mp+p+1)
2((m+p+p^2m^2-pm)=m^2p+1+mp^2+mp
m(2-mp)+p(2-mp)+mp((2mp-1)-(2mp+1)
put mp=r
(m+p)(2-r)+r(2r-1)-(2r+1)
put m+p=y
2r^2-r(3+y)+2y+1
solving for r
[(3+y)+/-root(1+y^2-10y)]/4
considering only the positive value(just for simplicity)
[(3+y)+root(1+y^2-10y)]/4=r............................................j
we know that m+p=y
mp=r
(m-p)^2=(m+p)^2-mp
(m-p)=root(y^2-4r)
m=[root(y^2-4r)+y]/2..........................................................f
p=2r/[root(y^2-4r)+y] (as mp=r)...........................................1
but from the equation p=[y-root(y^2-4r)]/2..................................g
from f
2m-y=[root(y^2-4r)
(2m-y)^2=(y^2-4r)
from g
2p-y=-root(y^2-4r)
(2p-y)^2=(y^2-4r)
from these two results
p=m is one of the soln
equating 1 and f
2r/[root(y^2-4r)+y] =[root(y^2-4r)+y]/2
4r=y^2-4r+y^2+2y(root(y^2-4r)
4r-y^2=y(root(y^2-4r)
squaring on both sides
16r^2+y^4-8ry^2=y^2(y^2-4r)
16r^2+y^4-8ry^2=y^4-4ry^2
4r=y^2
substituting the value or r from j
(3+y)+root(1+y^2-10y)=y^2
3+y-y^2=-root(1+y^2-10y)
9+y^4+6y-2y^3-6y^2=1-10y
y^4-2y^3-6y^2+16y+8=0
let t,y,u,i be the solutions of y(like i am not getting nice integers after trying with calculator)
we know that y=p+m
and that p=m
p=m=y/2
p=c/b and m=b/a
as p=m , b^2=ac
also mp=r=c/a
p^2=m^2=c/a
let p=m=i/2(putting one of the soln of y)
i^2/4=c/a
4c=ai^2
we know that b^2=ac
b^2=aai^2/4
b=ai/2(considering only the positive value)
so the sides are related as a,b,c=a,ai/2,ai^2/4, it can be an equilateral triangle if i=2
but putting the value of i in y^4-2y^3-6y^2+16y+8=0, it is not coming out to be 0
for this to be a right angled triangle
a^2i^2/4=a^2i^4/16+a^2
putting i^2/4=h
h=h^2+1
solving, we get imaginary roots, so this is not a rt angled triangle too
it is not coming ut to be 0 , so it is not a rt angled triangle
for it to be isosceles, any of these three conditions should be satisfied a=ai/2,ai/2=ai^2/4,a=ai^2/4
let us take the first condition a=ai/2
i=2 is not the soln
so , it is rooted out
ai/2=ai^2/4
it is also rooted out
so, we are left with only
a=ai^2/4
i=+/-2
substituting y=-2 in
y^4-2y^3-6y^2+16y+8=0
it does not satisfy the equation
hence , it is not an isosceles nor equilateral triangle nor a rt triangle
the value of y is nearly equal to 0.45
i will post the soln tomorrow