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Question
Prove that `(x + y + z)/((xy)^-1 + (yz)^-1 + (zx)^-1) = xyz`.
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Solution
Given:
Variables x, y, z are all non-zero and expression to analyze:
`(x + y + z)/((xy)^-1 + (yz)^-1 + (zx)^-1)`
To Prove:
`(x + y + z)/((xy)^-1 + (yz)^-1 + (zx)^-1) = xyz`
Proof:
Step 1: Write the denominator with clearer terms.
`(xy)^-1 = 1/(xy)`
`(yz)^-1 = 1/(yz)`
`(zx)^-1 = 1/(zx)`
So the denominator becomes:
`1/(xy) + 1/(yz) + 1/(zx)`
Step 2: Find the common denominator of the denominator terms.
Common denominator is (xyz).
Express each term:
`1/(xy) = z/(xyz)`
`1/(yz) = x/(xyz)`
`1/(zx) = y/(xyz)`
So, `1/(xy) + 1/(yz) + 1/(zx) = (z + x + y)/(xyz)`.
Step 3: Substitute the denominator back into the main expression:
`(x + y + z)/((z + x + y)/(xyz)) = (x + y + z) xx (xyz)/(z + x + y)`
Step 4: Recognize that the numerator and denominator inside the fraction are the same addition is commutative, so (x + y + z = z + x + y)
`xyz xx (x + y + z)/(x + y + z) = xyz`
Thus, `(x + y + z)/((xy)^-1 + (yz)^-1 + (zx)^-1) = xyz` is proved.
