2
$\begingroup$

Please help me prove the equality: If $xyz=1$, prove that

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

$\endgroup$
7
$\begingroup$

Note that $$\dfrac{x}{xy+x+1} = \dfrac{xz}{xyz+xz + z} = \dfrac{xz}{xz+z+1}$$ Similarly, we get $$\dfrac{y}{yz+y+1} = \dfrac1{z+1+\dfrac1y} = \dfrac1{z+1+xz}$$ Now add the three terms and finish it off by canceling the numerator and denominator.

$\endgroup$
4
$\begingroup$

Hint: Rewrite the $1$ in the first denominator as $xyz,$ so that the left hand side becomes $$\begin{align}\frac1{y+1+yz}+\frac{y}{yz+y+1}+\frac{z}{xz+z+1} &= \frac{1+y}{yz+y+1}+\frac{z}{xz+z+1}\\ &= 1-\frac{yz}{yz+y+1}+\frac{z}{xz+z+1}.\end{align}$$ Can you go from here?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.