a)[tex]A=\frac{a^2+ab}{a+2b}+\frac{b^2+ab}{2a+b}=\frac{a(a+b)}{a+2b}+\frac{b(a+b)}{2a+b}=(a+b)(\frac{a}{a+2b}+\frac{b}{2a+b})=(a+b)(\frac{2a+2b}{a+2b}+\frac{2a+2b}{2a+b}-2)=2(a+b)[(a+b)(\frac{1}{a+2b}+\frac{1}{b+2a})-1]\geq 2(a+b)[(a+b).\frac{4}{3(a+b)}-1]=2(a+b)(\frac{4}{3}-1)=\frac{2}{3}(a+b)[/tex]
Lại có;[tex]a+b+2ab=12\Rightarrow a+b=12-2ab\geq 12-2.(\frac{a+b}{2})^2=12-\frac{1}{2}(a+b)^2\Rightarrow \frac{1}{2}(a+b)^2+(a+b)-12\geq 0\Rightarrow (a+b)^2+2(a+b)-24\geq 0\Rightarrow [(a+b)-4][(a+b)+6]\geq 0\Rightarrow a+b\geq 4 hoặc a+b\leq -6(loại vì a,b\geq 0)[/tex]
[tex]\Rightarrow A\geq \frac{2}{3}(a+b)\geq \frac{8}{3}[/tex]