Given u = \sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } $$$$+ \sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } $\therefore$ ${u^2} = {a^2}{\cos ^2}\theta + {b^2}{\sin ^2}\theta + {a^2}{\sin ^2}\theta + {b^2}{\cos ^2}\theta$ $+ 2\sqrt {\left( {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } \right)} \times \sqrt {\left( {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } \right)}$ $\Rightarrow {u^2} =$ {a^2} + {b^2}$$$$ + $$$$2\sqrt {\left( {{a^4} + {b^4}} \right){{\cos }^2}\theta {{\sin }^2}\theta + {a^2}{b^2}\left( {{{\cos }^4}\theta + {{\sin }^4}\theta } \right)} $\Rightarrow {u^2} =$ {a^2} + {b^2}$$$$ + $$$$2\sqrt {\left( {{a^4} + {b^4}} \right){{\cos }^2}\theta {{\sin }^2}\theta + {a^2}{b^2}\left( {1 - 2{{\cos }^2}\theta {{\sin }^2}\theta } \right)} $\Rightarrow {u^2} =$ {a^2} + {b^2}$$$$ + $$$$2\sqrt {\left( {{a^4} + {b^4} - 2{a^2}{b^2}} \right){{\cos }^2}\theta {{\sin }^2}\theta + {a^2}{b^2}} $\Rightarrow {u^2} =$ {a^2} + {b^2}$$$$ + $$$$2\sqrt {{{\left( {{a^2} - {b^2}} \right)}^2}{{{{\left( {2\cos \theta \sin \theta } \right)}^2}} \over 4} + {a^2}{b^2}} $\Rightarrow {u^2} =$ {a^2} + {b^2}$$$$ + $$$$2\sqrt {{{\left( {{a^2} - {b^2}} \right)}^2}{{{{\sin }^2}2\theta } \over 4} + {a^2}{b^2}} We know $0 \le {\sin ^2}2\theta \le 1$ $\therefore$ $0 \le {\left( {{a^2} - {b^2}} \right)^2}{{{{\sin }^2}2\theta } \over 4} \le {{{{\left( {{a^2} - {b^2}} \right)}^2}} \over 4}$ $\Rightarrow$ ${a^2}{b^2} \le$ {\left( {{a^2} - {b^2}} \right)^2}{{{{\sin }^2}2\theta } \over 4} + {a^2}{b^2}$$$$ \le {{{{\left( {{a^2} - {b^2}} \right)}^2}} \over 4} + {a^2}{b^2} $\therefore$ Min value of ${u^2} = {a^2} + {b^2}$ $+ 2\sqrt {{a^2}{b^2}}$ = ${\left( {a + b} \right)^2}$ and Max value of ${u^2} = {a^2} + {b^2}$ $+ 2\sqrt {{{{{\left( {{a^2} - {b^2}} \right)}^2}} \over 4} + {a^2}{b^2}}$ $= {a^2} + {b^2}$ $+ 2\sqrt {{{{{\left( {{a^2} - {b^2}} \right)}^2} + 4{a^2}{b^2}} \over 4}}$ $= {a^2} + {b^2}$ $+ 2\sqrt {{{{{\left( {{a^2} + {b^2}} \right)}^2}} \over 4}}$ $= {a^2} + {b^2}$ $+\, {a^2} + {b^2}$ ${ = 2\left( {{a^2} + {b^2}} \right)}$ Max of ${u^2}$ - Min of ${u^2}$ = ${2\left( {{a^2} + {b^2}} \right)}$ - ${\left( {a + b} \right)^2}$ = ${2\left( {{a^2} + {b^2}} \right)}$ - ${\left( {{a^2} + {b^2} + 2ab} \right)}$ = $\sqrt {{{ = 2\left( {{a^2} + {b^2} - 2ab} \right)} \over 4}}$ = ${\left( {a - b} \right)^2}$