3)
Ta có: \[{{(ac+bd)}^{2}}+{{(ad-bc)}^{2}}={{a}^{2}}{{c}^{2}}+2abcd+{{b}^{2}}{{d}^{2}}+{{a}^{2}}{{d}^{2}}-2abcd+{{b}^{2}}{{c}^{2}}\]
\[={{a}^{2}}\left( {{c}^{2}}+{{d}^{2}} \right)+{{b}^{2}}\left( {{d}^{2}}+{{c}^{2}} \right)=\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{c}^{2}}+{{d}^{2}} \right)\]
Vì \[ad-bc=1\] nên \[1+{{(ac+bd)}^{^{2}}}=\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{c}^{2}}+{{d}^{2}} \right)\quad (1)\]
áp dụng bất đẳng thức Côsi cho hai số không âm \[\left( {{a}^{2}}+{{b}^{2}} \right)\quad ;\left( {{c}^{2}}+{{d}^{2}} \right)\] có: \[P={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}+ac+bd\ge 2\sqrt{\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{c}^{2}}+{{d}^{2}} \right)}+ac+bd\]
\[\Rightarrow P\ge 2\sqrt{1+{{\left( ac+bd \right)}^{2}}}+ac+bd\] (theo (1))
Rõ ràng \[P>0\] vì: \[2\sqrt{1+{{\left( ac+bd \right)}^{2}}}>{{\left| ac+bd \right|}^{2}}\]
Đặt \[x=ac+bd\],ta có: \[P\ge 2\sqrt{1+{{x}^{2}}}+x\]\[\Leftrightarrow {{P}^{2}}\ge 4\left( 1+{{x}^{2}} \right)+4x\sqrt{1+{{x}^{2}}}+{{x}^{2}}=\left( 1+{{x}^{2}} \right)+4x\sqrt{1+{{x}^{2}}}+4{{x}^{2}}+3\]
\[={{\left( \sqrt{1+{{x}^{2}}}+2x \right)}^{2}}+3\ge 3\]
Vậy \[P\ge \sqrt{3}\]