Question

Solve this problem....give entire process.​

Answer 1

$\green{\large\underline{\sf{Given- }}}$

If pqr = 1, prove that

$\sf{ \:\:\dfrac{1}{1 + p + {q}^{ - 1} } + \dfrac{1}{1 + q + {r}^{ - 1} } + \dfrac{1}{1 + r + {p}^{ - 1} }=1}$

$\green{\large\underline{\sf{Solution-}}}$

Given that,

$\green{\rm :\longmapsto\:pqr = 1}$

Consider,

$\blue{\bf :\longmapsto\:\dfrac{1}{1 + p + {q}^{ - 1} }}$

$\rm \:  =  \: \dfrac{1}{1 + p + \dfrac{1}{q} }$

$\rm \:  =  \: \dfrac{q}{q + pq + 1}$

$\blue{\bf :\longmapsto\:\dfrac{1}{1 + p + {q}^{ - 1} }= \dfrac{q}{pq + q + 1} }$

Now, Consider

$\blue{\bf :\longmapsto\:\dfrac{1}{1 + q + {r}^{ - 1} }}$

$\rm \:  =  \: \dfrac{1}{1 + q + pq}$

$\red{\bigg \{ \because \:pqr = 1 \: \: \rm \implies\:pq = {r}^{ - 1} \bigg \}}$

So,

$\blue{\bf :\longmapsto\:\dfrac{1}{1 + q + {r}^{ - 1} }= \dfrac{1}{pq + q + 1} }$

Now, Consider,

$\blue{\bf :\longmapsto\:\dfrac{1}{1 + r + {p}^{ - 1} }}$

$\rm \:  =  \: \dfrac{1}{1 + \dfrac{1}{pq} + \dfrac{1}{p} }$

$\rm \:  =  \: \dfrac{1}{\dfrac{pq + 1 + q}{pq} }$

$\rm \:  =  \: \dfrac{pq}{pq +q + 1}$

So,

$\blue{\bf :\longmapsto\:\dfrac{1}{1 + r + {p}^{ - 1} }= \dfrac{pq}{pq + q + 1} }$

Now, Consider,

$\blue{\bf :\longmapsto\:\dfrac{1}{1 + p + {q}^{ - 1} } + \dfrac{1}{1 + q + {r}^{ - 1} } + \dfrac{1}{1 + r + {p}^{ - 1} } }$

$\rm \:  =  \: \dfrac{q}{pq + q + 1} + \dfrac{1}{pq + q + 1} + \dfrac{pq}{pq + q + 1}$

$\rm \:  =  \: \dfrac{pq + q + 1}{pq + q + 1}$

$\rm \:  =  \: 1$

Hence,

$\red{ \boxed{ \sf{ \:\:\dfrac{1}{1 + p + {q}^{ - 1} } + \dfrac{1}{1 + q + {r}^{ - 1} } + \dfrac{1}{1 + r + {p}^{ - 1} } = 1}}}$