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If $x+y+z=xyz$, prove that: $$\frac {x}{1-x^2} +\frac {y}{1-y^2} + \frac {z}{1-z^2}=\frac {4xyz}{(1-x^2)(1-y^2)(1-z^2)}$$.

My Attempt:

$$L.H.S=\frac {x}{1-x^2}+\frac {y}{1-y^2}+\frac {z}{1-z^2}$$ $$=\frac {x(1-y^2)(1-z^2)+y(1-x^2)(1-z^2)+z(1-x^2)(1-y^2)}{(1-x^2)(1-y^2)(1-z^2)}$$ $$=\frac {x+y+z-xz^2-xy^2+xy^2z^2-yz^2-yx^2+x^2yz^2-zy^2-zx^2+zx^2y^2}{(1-x^2)(1-y^2)(1-z^2)}$$.

I could not move on from here.Please help.

Thanks

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    Put $x=\tan A,y=\tan B,z=\tan C$ – 2017-02-01
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    @DURGESHTIWARI You would have to prove that all such $x,y,z$ can be expressed in a form regarding $\tan A, \tan B, \tan C$ assuming $A+B+C=\pi$. It is known that if $x=\tan A, y=\tan B, z=\tan C$, then $$x+y+z=xyz$$ But you would ahve to proze the reverse holds. – 2017-02-01
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    https://math-kali.blogspot.in/2010/07/if-x-y-z-xyz.html OR https://books.google.co.in/books?id=UwLoDQAAQBAJ&pg=PA189&lpg=PA189&dq=x+y+z+xyz++tan++4xyz&source=bl&ots=xnCyj34r1I&sig=9AYNIuCgX3t0-2KuQzKq1sH12lQ&hl=en&sa=X&ved=0ahUKEwi4_c-6ou_RAhXJv48KHZXsDM4Q6AEIKzAF#v=onepage&q=x%20y%20z%20xyz%20%20tan%20%204xyz&f=false – 2017-02-01

2 Answers 2

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Continuing from where you left, expressing terms of the numerator as: $$-xz^2-x^2z+x^2yz^2 =-xz (z+x-xyz) =-xz (-y) $$ $$-xy^2-x^2y+x^2y^2z =-xy (x+y-xyz) =-xy (-z) $$ $$-yz^2-y^2z+xy^2z^2 =-yz (y+z-xyz)=-yz (-x) $$ $$x+y+z=xyz $$

Now add everything and the result follows. Hope it helps.

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    @@Rohan, Are there other alternatives, too? – 2017-02-01
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$\textbf{HINT:}$ Try putting $x=tan(\alpha);y=tan(\beta);z=tan(\gamma)$ Use: $$tan(2\theta)=\frac{2tan(\theta)}{1-tan^2(\theta)}$$ $\textbf{Note that $\alpha +\beta+\gamma=\pi$}$ by the condition since: $$tan(\alpha+\beta+\gamma)=\frac{\Sigma tan(\alpha)-tan(\alpha)tan(\beta)tan(\gamma)}{1-\Sigma tan(\alpha)tan(\beta)}$$ Where $\Sigma$ denotes cyclic summation.