Solve Algebra Challenge: $(x+1)(y+1)/(x+y)+\cdots

[anemone]

Given that $x,\,y$ and $z$ are non-zero real numbers such that $x + y + z = 3$ and $xy + yz + zx = −1$.

Evaluate \(\displaystyle \frac{(x + 1)(y + 1)}{x + y}+ \frac{(y + 1)(z + 1)}{y + z}+ \frac{(z + 1)(x + 1)}{z + x}\).

 

[Euge]

Hi anemone,

Here is my solution.

Spoiler

The expression reduces to zero. Using the cyclic summation notation $\sum_\sigma$, we write the expression as

$$\sum_\sigma \frac{(x+1)(y+1)}{x + y}.$$

Note $(x + 1)(y + 1) = (x + y) + (xy + 1) = (x + y) - (yz + zx) = (x + y)(1 - z)$. Thus

$$\sum_\sigma \frac{(x + 1)(y + 1)}{x + y} = \sum_\sigma (1 - z) = 3 - \sum_\sigma z = 0.$$

 

[Greg]

Given that $x,\,y$ and $z$ are non-zero real numbers such that $x + y + z = 3$ and $xy + yz + zx = −1$.

Evaluate \(\displaystyle \frac{(x + 1)(y + 1)}{x + y}+ \frac{(y + 1)(z + 1)}{y + z}+ \frac{(z + 1)(x + 1)}{z + x}\).

Spoiler

$$\frac{(x+1)(y+1)}{x+y}=\frac{xy+x+y+1}{x+y}=\frac{-1-z(3-z)+3-z+1}{3-z}=1-z$$

Similarly for the other two summands:

$$1-z+1-y+1-x=3-3=0$$

 

[anemone]

Very well done to both of you! (Cool)