Prove that \(\frac{1}{1+p+q^{-1}}+\frac{1}{1+q+r^{-1}}+\frac{1}{1+r+p^{-1}}=1\), if \(pqr=1\)

Solution: Given: \(pqr=1\) To prove: \(\frac{1}{1+p+q^{-1}}+\frac{1}{1+q+r^{-1}}+\frac{1}{1+r+p^{-1}}=1\) \(\text{L.H.S}=\frac{1}{1+p+q^{-1}}+\frac{1}{1+q+r^{-1}}+\frac{1}{1+r+p^{-1}}\) We will make the denominator of each term same by using the given condition \(pqr=1\). \(q^{-1}=\frac{1}{q}\)         \([\, ∵ a^{-1}=\frac{1}{a}]\) \(pqr=1\implies pq=\frac{1}{r}=r^{-1}\)  \([\, ∵ a^{-1}=\frac{1}{a}]\) \(pqr=1\implies r=\frac{1}{pq} \) Substituting these values, we get \(\text{L.H.S}= \frac{1}{1+p+\frac{1}{q}}+\frac{1}{1+q+pq}+\frac{1}{1+\frac{1}{pq}+\frac{1}{p}}\) \(= \frac{1}{\frac{q+pq+1}{q}}+\frac{1}{1+q+pq}+\frac{1}{\frac{pq+1+q}{pq}}\) \(=\frac{q}{q+pq+1}+\frac{1}{1+q+pq}+\frac{pq}{pq+1+q}\) \(=\frac{q}{1+q+pq}+\frac{1}{1+q+pq}+\frac{pq}{1+q+pq}\) \(=\frac{q+1+pq}{1+q+pq}\) \(=\frac{1+q+pq}{1+q+pq}\) \(=1=\text{R.H.S}.\) Therefore, if \(\mathbf{pqr=1}\), then \(\mathbf{\frac{1}{1+p+q^{-1}}+\frac{1}{1+q+r^{-1}}+\frac{1}{1+r+p^{-1}}=1}.\)  

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Alternative Method: Given: \(pqr=1\), this equation is satisfied for \(p=q=r=1\). To prove: \(\frac{1}{1+p+q^{-1}}+\frac{1}{1+q+r^{-1}}+\frac{1}{1+r+p^{-1}}=1\) \(\text{L.H.S}=\frac{1}{1+p+q^{-1}}+\frac{1}{1+q+r^{-1}}+\frac{1}{1+r+p^{-1}}\) Substituting the values of \(p=q=r=1\), we get \(\text{L.H.S}=\frac{1}{1+1+1^{-1}}+\frac{1}{1+1+1^{-1}}+\frac{1}{1+1+1^{-1}}\) \(=\frac{1}{1+1+1}+\frac{1}{1+1+1}+\frac{1}{1+1+1}\) \(=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\) \(=\frac{1+1+1}{3}\) \(=\frac{3}{3}\) \(=1=\text{R.H.S}\) Therefore, if \(\mathbf{pqr=1}\), then \(\mathbf{\frac{1}{1+p+q^{-1}}+\frac{1}{1+q+r^{-1}}+\frac{1}{1+r+p^{-1}}=1}.\)