- Nếu \(ab=0\Rightarrow M=4+2\sqrt{2}\) (1)
- Nếu \(ab\ne0\)
\(M=\frac{\frac{a^4}{a}+\frac{b^4}{b}+4}{ab+1}\ge\frac{\frac{\left(a^2+b^2\right)^2}{a+b}+4}{\frac{a^2+b^2}{2}+1}\ge\frac{\frac{\left(a^2+b^2\right)^2}{\sqrt{2\left(a^2+b^2\right)}}+4}{\frac{a^2+b^2}{2}+1}=3\) (2)
So sánh (1) và (2) \(\Rightarrow M_{min}=3\) khi \(a=b=1\)
- Do \(a^2+b^2=2\Rightarrow0\le a;b\le\sqrt{2}\)
\(\Rightarrow a\left(a-\sqrt{2}\right)\le0\Rightarrow a^2\le a\sqrt{2}\Rightarrow a^3\le a^2\sqrt{2}\)
Tương tự \(b^3\le b^2\sqrt{2}\) \(\Rightarrow a^3+b^3\le\left(a^2+b^2\right)\sqrt{2}=2\sqrt{2}\)
\(\Rightarrow M=\frac{a^3+b^3+4}{ab+1}\le\frac{4+2\sqrt{2}}{ab+1}\le4+2\sqrt{2}\)
\(\Rightarrow M_{max}=4+2\sqrt{2}\) khi \(\left(a;b\right)=\left(0;\sqrt{2}\right);\left(\sqrt{2};0\right)\)
