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 \(\mathbf{\frac{1}{1+p+q^{-1}}+\frac{1}{1+q+r^{-1}}+\frac{1}{1+r+p^{-1}}=1}.\)

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 \(\mathbf{\frac{1}{1+p+q^{-1}}+\frac{1}{1+q+r^{-1}}+\frac{1}{1+r+p^{-1}}=1}.\)