Question

x=a(y+z), y=b(z+x), z=c(x+y), a, b, c
is constant then proved 2abc+ ab+ bc+ ca =1​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:x = a(y + z)$

$\bf\implies \:a = \dfrac{x}{y + z}$

Also,

$\rm :\longmapsto\:y = b(z + x)$

$\bf\implies \:b = \dfrac{y}{z + x}$

Also,

$\rm :\longmapsto\:z = c(x + y)$

$\bf\implies \:c = \dfrac{z}{x + y}$

Now, Consider,

$\red{\rm :\longmapsto\:ab + bc + ca}$

$\rm \:  =  \: \dfrac{xy}{(y + z)(z + x)} + \dfrac{yz}{(z + x)(x + y)} + \dfrac{zx}{(x + y)(y + z)}$

$\rm \:  =  \: \dfrac{xy(x + y) + yz(y + z) + zx(z + x)}{(x + y)(y + z)(z + x)}$

Now,

Consider

$\red{\rm :\longmapsto\:2abc + ab + bc + ca}$

On substituting the values, we get

$\rm =\dfrac{2xyz}{(x + y)(y + z)(z + x)} + \dfrac{xy(x + y) + yz(y + z) + zx(z + x)}{(x + y)(y + z)(z + x)}$

$\rm = \dfrac{2xyz + xy(x + y) + yz(y + z) + zx(z + x)}{(x + y)(y + z)(z + x)}$

$\rm = \dfrac{2xyz + {yx}^{2} + {xy}^{2} + {zy}^{2} + {yz}^{2} + {zx}^{2} + {xz}^{2} }{(x + y)(y + z)(z + x)}$

on rearranging the terms, we have

$\rm \:  = \dfrac{( {yx}^{2}+{yz}^{2} + 2xyz) + ( {xy}^{2} + {zy}^{2}) + ( {zx}^{2} + {xz}^{2}}{(x + y)(y + z)(z + x)}$

$\rm \:  = \dfrac{y({x}^{2}+{z}^{2} + 2xz) + {y}^{2}(x + z) +zx(x + z)}{(x + y)(y + z)(z + x)}$

$\rm \:  = \dfrac{y {(x + z)}^{2} + {y}^{2}(x + z) +zx(x + z)}{(x + y)(y + z)(z + x)}$

$\rm \:  =  \: \dfrac{\cancel{(z + x)} \: \: [y(z + x) + {y}^{2} + zx]}{(x + y)(y + z) \: \cancel{(z + x)}}$

$\rm \:  =  \: \dfrac{y(z + x) + {y}^{2} + zx}{(x + y)(y + z) }$

$\rm \:  =  \: \dfrac{yz + xy+ {y}^{2} + zx}{(x + y)(y + z) }$

On rearranging the terms, we get

$\rm \:  =  \: \dfrac{(yz + zx)+ ({y}^{2} + yx)}{(x + y)(y + z) }$

$\rm \:  =  \: \dfrac{z(y + x)+ y({y} + x)}{(x + y)(y + z) }$

$\rm \:  =  \: \dfrac{(z + y)(y + x)}{(x + y)(y + z) }$

$\rm \:  =  \:\cancel{ \dfrac{(x + y)(y + z)}{(x + y)(y + z) } }$

$\rm \:  =  \: 1$

Hence,

$\red{\rm :\longmapsto\:2abc + ab + bc + ca = 1}$

Answer 2

Answer:

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:x = a(y + z)$

$\bf\implies \:a = \dfrac{x}{y + z}$

Also,

$\rm :\longmapsto\:y = b(z + x)$

$\bf\implies \:b = \dfrac{y}{z + x}$

Also,

$\rm :\longmapsto\:z = c(x + y)$

$\bf\implies \:c = \dfrac{z}{x + y}$

Now, Consider,

$\red{\rm :\longmapsto\:ab + bc + ca}$

$\rm \:  =  \: \dfrac{xy}{(y + z)(z + x)} + \dfrac{yz}{(z + x)(x + y)} + \dfrac{zx}{(x + y)(y + z)}$

$\rm \:  =  \: \dfrac{xy(x + y) + yz(y + z) + zx(z + x)}{(x + y)(y + z)(z + x)}$

Now,

Consider

$\red{\rm :\longmapsto\:2abc + ab + bc + ca}$

On substituting the values, we get

$\rm =\dfrac{2xyz}{(x + y)(y + z)(z + x)} + \dfrac{xy(x + y) + yz(y + z) + zx(z + x)}{(x + y)(y + z)(z + x)}$

$\rm = \dfrac{2xyz + xy(x + y) + yz(y + z) + zx(z + x)}{(x + y)(y + z)(z + x)}$

$\rm = \dfrac{2xyz + {yx}^{2} + {xy}^{2} + {zy}^{2} + {yz}^{2} + {zx}^{2} + {xz}^{2} }{(x + y)(y + z)(z + x)}$

on rearranging the terms, we have

$\rm \:  = \dfrac{( {yx}^{2}+{yz}^{2} + 2xyz) + ( {xy}^{2} + {zy}^{2}) + ( {zx}^{2} + {xz}^{2}}{(x + y)(y + z)(z + x)}$

$\rm \:  = \dfrac{y({x}^{2}+{z}^{2} + 2xz) + {y}^{2}(x + z) +zx(x + z)}{(x + y)(y + z)(z + x)}$

$\rm \:  = \dfrac{y {(x + z)}^{2} + {y}^{2}(x + z) +zx(x + z)}{(x + y)(y + z)(z + x)}$

$\rm \:  =  \: \dfrac{\cancel{(z + x)} \: \: [y(z + x) + {y}^{2} + zx]}{(x + y)(y + z) \: \cancel{(z + x)}}$

$\rm \:  =  \: \dfrac{y(z + x) + {y}^{2} + zx}{(x + y)(y + z) }$

$\rm \:  =  \: \dfrac{yz + xy+ {y}^{2} + zx}{(x + y)(y + z) }$

On rearranging the terms, we get

$\rm \:  =  \: \dfrac{(yz + zx)+ ({y}^{2} + yx)}{(x + y)(y + z) }$

$\rm \:  =  \: \dfrac{z(y + x)+ y({y} + x)}{(x + y)(y + z) }$

$\rm \:  =  \: \dfrac{(z + y)(y + x)}{(x + y)(y + z) }$

$\rm \:  =  \:\cancel{ \dfrac{(x + y)(y + z)}{(x + y)(y + z) } }$

$\rm \:  =  \: 1$

Hence,

$\red{\rm :\longmapsto\:2abc + ab + bc + ca = 1}$