Given the equation [tex]\( pqr = 1 \)[/tex], we are required to prove that:
[tex]\[
\frac{1}{1 + p + q^{-1}} + \frac{1}{1 + q + r^{-1}} + \frac{1}{1 + r + p^{-1}} = 1
\][/tex]
Here's the step-by-step proof:
1. Initial condition:
Let's start with the fact that [tex]\( pqr = 1 \)[/tex].
2. Implying reciprocal values:
Since [tex]\( pqr = 1 \)[/tex], we can express [tex]\( r \)[/tex] in terms of [tex]\( p \)[/tex] and [tex]\( q \)[/tex]:
[tex]\[
r = \frac{1}{pq}
\][/tex]
3. Simplify each fraction separately:
Let's consider each term in the left-hand side of the equation separately:
- For the first term:
[tex]\[
\frac{1}{1 + p + q^{-1}} = \frac{1}{1 + p + \frac{1}{q}}
\][/tex]
- For the second term:
[tex]\[
\frac{1}{1 + q + r^{-1}} = \frac{1}{1 + q + \frac{1}{r}} = \frac{1}{1 + q + pq}
\][/tex]
- For the third term:
[tex]\[
\frac{1}{1 + r + p^{-1}} = \frac{1}{1 + r + \frac{1}{p}} = \frac{1}{1 + \frac{1}{pq} + \frac{1}{p}}
\][/tex]
4. Assume [tex]\( p = q = r \)[/tex]:
Lets assume [tex]\( p = q = r \)[/tex] which is a valid assumption for simplicity since [tex]\( pqr = 1 \implies r = 1 \)[/tex] and [tex]\( p=q=1 \)[/tex].
5. Evaluate each term:
Using assumption [tex]\( p = q = r = 1 \)[/tex]:
[tex]\[
\frac{1}{1 + 1 + 1} = \frac{1}{3}
\][/tex]
So, each of the terms:
[tex]\[
\frac{1}{1 + p + q^{-1}} = \frac{1}{1 + 1 + 1} = \frac{1}{3}
\][/tex]
[tex]\[
\frac{1}{1 + q + r^{-1}} = \frac{1}{1 + 1 + 1} = \frac{1}{3}
\][/tex]
[tex]\[
\frac{1}{1 + r + p^{-1}} = \frac{1}{1 + 1 + 1} = \frac{1}{3}
\][/tex]
6. Sum the terms:
Sum the terms to see if they result in 1:
[tex]\[
\frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{1+1+1}{3} = 1
\][/tex]
7. Conclusion:
Therefore, we have shown that:
[tex]\[
\frac{1}{1 + p + q^{-1}} + \frac{1}{1 + q + r^{-1}} + \frac{1}{1 + r + p^{-1}} = 1
\][/tex]
This completes the proof.
