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

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The discussion focuses on evaluating the expression $$\frac{(x + 1)(y + 1)}{x + y}+ \frac{(y + 1)(z + 1)}{y + z}+ \frac{(z + 1)(x + 1)}{z + x}$$ under the conditions that $x + y + z = 3$ and $xy + yz + zx = -1$ for non-zero real numbers $x, y, z$. Participants share their solutions and methods for simplifying the expression. The conversation highlights the importance of substituting the given conditions into the expression to find a solution. The collaborative nature of the discussion encourages problem-solving and peer support among participants. Ultimately, the evaluation of the expression is central to the challenge presented.
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Given that $x,\,y$ and $z$ are non-zero real numbers such that $x + y + z = 3$ and $xy + yz + zx = −1$.

Evaluate $$\frac{(x + 1)(y + 1)}{x + y}+ \frac{(y + 1)(z + 1)}{y + z}+ \frac{(z + 1)(x + 1)}{z + x}$$.
 
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Hi anemone,

Here is my solution.

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

$$\sum_\sigma \frac{(x + 1)(y + 1)}{x + y} = \sum_\sigma (1 - z) = 3 - \sum_\sigma z = 0.$$
 
anemone said:
Given that $x,\,y$ and $z$ are non-zero real numbers such that $x + y + z = 3$ and $xy + yz + zx = −1$.

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

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

Similarly for the other two summands:

$$1-z+1-y+1-x=3-3=0$$
 
Very well done to both of you! (Cool)
 
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