\(\sqrt{3b\left(a+2b\right)}\le\frac{3b+\left(a+2b\right)}{2}\); \(\sqrt{3a\left(b+2a\right)}\le\frac{3a+\left(b+2a\right)}{2}\)
=> M\(\le a\frac{a+5b}{2}+b\frac{5a+b}{2}\)=\(\frac{a^2+b^2+10ab}{2}\)\(\le\frac{6\left(a^2+b^2\right)}{2}\)( áp dụng 2ab\(\le a^2+b^2\))=3(a2+b2)\(\le\)6
dấu = khi a =b =1