Question

If $lmn=1$, show that
${\displaystyle \frac{1}{1+l+m^{-1}}}+{\displaystyle \frac{1}{1+m+n^{-1}}}+{\displaystyle \frac{1}{1+n+l^{-1}}}=1$

Answer 1

Hint- Simplify the L.H.S and reduce it to a single term and try to cancel out numerator and denominator which gives unit value.
Given: $lmn=1$
To prove: ${\displaystyle \frac{1}{1+l+m^{-1}}}+{\displaystyle \frac{1}{1+m+n^{-1}}}+{\displaystyle \frac{1}{1+n+l^{-1}}}=1$
Taking first term, $${\displaystyle \frac{1}{1+l+m^{-1}}}$$
$$\begin{gathered} ={\displaystyle \frac{1}{1+l+{\displaystyle \frac{1}{m}}}} \\ ={\displaystyle \frac{m}{m+lm+1}} \end{gathered}$$
Taking second term, $${\displaystyle \frac{1}{1+m+n^{-1}}}$$
$$\begin{gathered} ={\displaystyle \frac{1}{1+m+{\displaystyle \frac{1}{n}}}} \\ ={\displaystyle \frac{1}{1+m+lm}}\left\{\because lmn=1\Rightarrow {\displaystyle \frac{1}{n}}=lm\right\} \end{gathered}$$
Taking third term, $${\displaystyle \frac{1}{1+n+l^{-1}}}$$
Multiply and divide by $lm$, we get:
$$\begin{gathered} ={\displaystyle \frac{1}{1+n+l^{-1}}}\times {\displaystyle \frac{lm}{lm}} \\ ={\displaystyle \frac{lm}{lm+lmn+l^{-1}lm}} \\ ={\displaystyle \frac{lm}{lm+1+m}}\left\{\because lmn=1\right\} \end{gathered}$$
Now, combining all these terms to form the L.H.S, we get:
$$\begin{gathered} \text{L}\text{.H}\text{.S}={\displaystyle \frac{1}{1+l+m^{-1}}}+{\displaystyle \frac{1}{1+m+n^{-1}}}+{\displaystyle \frac{1}{1+n+l^{-1}}} \\ ={\displaystyle \frac{1}{1+l+{\displaystyle \frac{1}{m}}}}+{\displaystyle \frac{1}{1+m+{\displaystyle \frac{1}{n}}}}+{\displaystyle \frac{1}{1+n+{\displaystyle \frac{1}{l}}}} \\ ={\displaystyle \frac{m}{m+lm+1}}+{\displaystyle \frac{1}{1+m+lm}}+{\displaystyle \frac{lm}{lm+1+m}} \end{gathered}$$
Since, these have common denominator, hence we can add them directly.
$$\begin{gathered} ={\displaystyle \frac{m+1+lm}{m+lm+1}} \\ =1 \\ =\text{R}\text{.H}\text{.S} \end{gathered}$$
Hence Proved.

Note- Whenever you see equations like these, always try to look for patterns to reduce the fraction and try to make denominators of each term equal in order to add the terms easily.