For two positive real numbers a and b such that ${1 \over {{a^2}}} + {1 \over {{b^3}}} = 4$, then minimum value of the constant term in the expansion of ${\left( {a{x^{{1 \over 8}}} + b{x^{ - {1 \over {12}}}}} \right)^{10}}$ is :

<p>Given, Binomial expansion,</p> <p>${\left( {a{x^{{1 \over 8}}} + b{x^{ - {1 \over {12}}}}} \right)^{10}}$</p> <p>General term,</p> <p>$${T_{r + 1}} = {}^{10}{C_r}\,.\,{\left( {a{x^{{1 \over 8}}}} \right)^{10 - r}}\,.\,{\left( {b{x^{ - {1 \over {12}}}}} \right)^r}$$</p> <p>$$ = {}^{10}{C_r}\,.\,{a^{10 - r}}\,.\,{b^r}\,.\,{x^{{{10 - r} \over 8}}}\,.\,{x^{ - {r \over {12}}}}$$</p> <p>$$ = {}^{10}{C_r}\,.\,{a^{10 - r}}\,.\,{b^r}\,.\,{x^{\left( {{{10 - r} \over 8} - {r \over {12}}} \right)}}$$</p> <p>$= {}^{10}{C_r}\,.\,{a^{10 - r}}\,.\,{b^r}\,.\,{x^{{{30 - 3r - 2r} \over {24}}}}$</p> <p>$= {}^{10}{C_r}\,.\,{a^{10 - r}}\,.\,{b^r}\,.\,{x^{{{30 - 5r} \over {24}}}}$</p> <p>For constant term,</p> <p>${{30 - 5r} \over {24}} = 0$</p> <p>$\Rightarrow r = 6$</p> <p>$\therefore$ Constant term,</p> <p>${T_{r + 1}} = {T_{6 + 1}} = {}^{10}{C_6}\,.\,{a^4}\,.\,{b^6}$</p> <p>$= {{10!} \over {6!\,4!}}{a^4}\,.\,{b^6}$</p> <p>$$ = {{10 \times 9 \times 8 \times 7} \over {4 \times 3 \times 2 \times 1}}\,.\,{a^4}\,.\,{b^6}$$</p> <p>$= 210{a^4}{b^6}$</p> <p>We know, $GM \ge HM$</p> <p>For terms a² and b³,</p> <p>$\sqrt {{a^2}{b^3}} \ge {2 \over {{1 \over {{a^2}}} + {1 \over {{b^3}}}}}$</p> <p>$\Rightarrow \sqrt {{a^2}{b^3}} \ge {2 \over 4}$</p> <p>$\Rightarrow {a^2}{b^3} \ge {1 \over 4}$</p> <p>$\Rightarrow {({a^2}{b^3})^2} \ge {1 \over {16}}$</p> <p>$\therefore$ ${a^4}{b^6} \ge {1 \over {16}}$</p> <p>$\therefore$ Minimum value of ${a^4}{b^6} = {1 \over {16}}$</p> <p>$\therefore$ Minimum value of constant term</p> <p>${T_7} = 210 \times {a^4}{b^6}$</p> <p>$= 210 \times {1 \over {16}}$</p> <p>$= {{105} \over 8}$</p>