JEE Advance - Mathematics (2000 - No. 14)
For every possitive integer $$n$$, prove that
$$\sqrt {\left( {4n + 1} \right)} < \sqrt n + \sqrt {n + 1} < \sqrt {4n + 2}.$$
Hence or otherwise, prove that $$\left[ {\sqrt n + \sqrt {\left( {n + 1} \right)} } \right] = \left[ {\sqrt {4n + 1} \,\,} \right],$$
where $$\left[ x \right]$$ denotes the gratest integer not exceeding $$x$$.
$$\sqrt {\left( {4n + 1} \right)} < \sqrt n + \sqrt {n + 1} < \sqrt {4n + 2}.$$
Hence or otherwise, prove that $$\left[ {\sqrt n + \sqrt {\left( {n + 1} \right)} } \right] = \left[ {\sqrt {4n + 1} \,\,} \right],$$
where $$\left[ x \right]$$ denotes the gratest integer not exceeding $$x$$.
The given inequality directly proves the required equality of the greatest integer functions.
Squaring all terms in the inequality is a key step to remove square roots and simplify the expressions, facilitating a comparison of integer parts.
The left inequality, \(\sqrt {4n + 1} < \sqrt n + \sqrt {n + 1} \), is sufficient to prove the equality of the greatest integer functions.
The right inequality, \(\sqrt n + \sqrt {n + 1} < \sqrt {4n + 2} \), is sufficient to prove the equality of the greatest integer functions.
The problem cannot be solved using the given inequalities.