[tex](a+b)(c+d)=2=ac+ad+bc+bd\\(a+c)(b+d)=3=ab+ad+cb+cd\\3-2=1=ab-ac-db+dc\\\Rightarrow 5=ab-ac-bd+dc+(a+d)(b+c)\\=2(ab+cd)[/tex]
Ta có
[tex]a^2+b^2+c^2+d^2\geq 2ab+2cd=5[/tex]
[tex](a+b)(c+d)=2=ac+ad+bc+bd\\(a+c)(b+d)=3=ab+ad+cb+cd\\3-2=1=ab-ac-db+dc\\\Rightarrow 5=ab-ac-bd+dc+(a+d)(b+c)\\=2(ab+cd)[/tex]
Ta có
[tex]a^2+b^2+c^2+d^2\geq 2ab+2cd=5[/tex]