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If $x.y.z=1$ show that $\dfrac 1{1+x+y^{-1}}+\dfrac1{1+y+z^{-1}}+\dfrac1{1+z+x^{-1}}=1$

My attempt - L.H.S$=\dfrac 1{1+x+y^{-1}}+\dfrac1{1+y+z^{-1}}+\dfrac1{1+z+x^{-1}}$

$=\dfrac y{y+xy+1}+\dfrac z{z+yz+1}+\dfrac x{x+zx+1}$

$=\dfrac{yz}{yz+xyz+z}+\dfrac z{z+yz+1}+\dfrac x{x+zx+1}$

What to do next?

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$$\frac1{1+x+y^{-1}}=\frac y{y+xy+1}=\frac{yz}{yz+xyz+z}(\text{ multiplying by } z)$$

$$\implies\frac1{1+x+y^{-1}}=\frac{yz}{yz+1+z}$$

$$\frac1{1+y+z^{-1}}=\frac z{z+yz+1}(\text{ multiplying by } z)$$

$$\frac1{1+z+x^{-1}}=\frac x{x+zx+1}=\frac{xyz}{xyz+yz(zx)+yz}(\text{ multiplying by } yz)$$

$$\implies\frac1{1+z+x^{-1}}=\frac1{1+z+yz}$$

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  • $\begingroup$ once again thanks a lot. $\endgroup$ – SNEHIL SINHA Sep 11 '14 at 10:00
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yours continue

$$\frac{yz}{yz+xyz+z}+\frac z{z+yz+1}+\frac x{x+zx+1}$$

$$=\frac{yz}{yz+1+z}+\frac{xz}{xz+xyz+x}+\frac{x}{x+zx+1}$$

$$=\frac{xyz}{xyz+x+zx}+\frac{xz}{xz+1+x}+\frac{x}{x+zx+1}$$

$$=\frac{1}{1+x+zx}+\frac{xz}{xz+1+x}+\frac{x}{x+zx+1}$$

$$=\frac{1+x+zx}{1+x+zx}$$

$$=1$$

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