# if x=a(y+z), y= b(z+x) & z= c(x+y) then, prove that, ab+bc+ca+2abc=1. no

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### 3 Answers

`a = x/(y+z)`

`b = y/(z+x)`

`c = z/(x+y)`

`ab = (xy)/((y+z)(z+x))`

`bc = (yz)/((z+x)(x+y))`

`ca = (xz)/((y+z)(x+y))`

`abc = (xyz)/((y+z)(x+y)(x+z))`

`ab+bc+ca+2abc`

`=(xy)/((y+z)(z+x))+(yz)/((z+x)(x+y))+(xz)/((y+z)(x+y))+(2xyz)/((y+z)(x+y)(x+z))`

Now we get `(x+y)(y+z)(z+x)` as the common denominator.

`= ((xy(x+y)+yz(y+z)+xz(x+z)+2xyz))/((y+z)(x+y)(x+z))`

`= (x^2y+xy^2+y^2z+z^2y+x^2z+z^2x+2xyz)/((y+z)(x+y)(x+z))`

`= (x^2y+2xyz+z^2y+xy^2+zy^2+x^2z+z^2x)/((y+z)(x+y)(x+z))`

`= (y(x^2+2xz+z^2)+y^2(x+z)+zx(x+z))/((y+z)(x+y)(x+z))`

(x+z) term is common for all.

`= ((x+z)(y(x+z)+y^2+zx)/((y+z)(x+y)(x+z))`

`= ((x+z)(yx+yz+y^2+zx))/((y+z)(x+y)(x+z))`

`= ((x+z)(y(x+y)+z(x+y)))/((y+z)(x+y)(x+z))`

`= ((x+z)(x+y)(y+z))/((y+z)(x+y)(x+z))`

`= 1`

*So the answer is proved.*

Given : x=a(y+z), y=b(x+z) and z=c(x+y)

Therefore : a=x/(y+z), b=y/(x+z) and= c=z/(x+y)

Require to prove : ab+bc+ca+2abc =1

From given we find the value of ab, bc, ca and 2abc

ab=(x/(y+z))*(y/(x+z))=xy/(y+z)(x+z)

bc=(y/(x+z))*(z/(x+y))=yz/(x+z)(x+y)

ca=(z/(x+y))(x/(y+z))=xz/(x+y)(y+z)

2abc=2(x/(y+z)*(y/(x+z)*(z/(x+y)=(2xyz)/(y+z)(x+z)(x+y)

L.H.S :-

ab+bc+ca+2abc=

=xy/(y+z)(x+z) +yz/(x+z)(x+y) +xz/(x+y)(y+z) +(2xyz)/(y+z)(x+z)(x+y)

=(xy(x+y)+yz(y+z)+xz(x+z)+2xyz)/(y+z)(x+z)(x+y)

=(x^2.y+x.y^2+y^2.z+y.z^2+x^2.z+x.z^2+2xyz)/(y+z)(x+z)(x+y)

=(x^2.y+2xyz+y.z^2+x.y^2+y^2.z+x^2.z+x.z^2)/(y+z)(x+z)(x+y)

=(y(x^2+2xz+z^2)+y^2(x+z)+xz(x+z))/(y+z)(x+z)(x+y)

=(y(x+z)^2+y^2(x+z)+xz(x+z))/(y+z)(x+z)(x+Y)

=[(x+z)(y.(x+z))+y^2(x+z)+xz(x+z)]/(y+z)(x+z)(x+y)

=[(x+z){y.(x+z)+y^2+xz)}]/(y+z)(x+z)(x+y)

=[(x+z)(xy+yz+y^2+xz)]/(y+z)(x+z)(x+y)

=[(x+z){(xy+y^2)+(yz+xz)}]/(y+z)(x+z)(x+y)

=[(x+z){y(x+y)+z(x+y)}]/(y+z)(x+z)(x+y)

=[(x+z)(x+y)(y+z)]/(y+z)(x+z)(x+y)

= 1

= R.H.S

**Proved**

1.x = a(y+z)

then

a = x/(y+z)

similarly, solve for b and c. Then plug all values into

ab+bc+ca+2abc

and hopefully it will simplify to 1.