Ta có: \(ab\le\frac{\left(a+b\right)^2}{4}\Rightarrow\left(a+b\right)^2\ge4\Rightarrow a+b\ge2\)
Và \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\ge ab\left(a+b\right)\ge2\)
Áp dụng BĐT AM-GM ta có:
\(\frac{a^3}{b+1}=a^3-\frac{a^3b}{b+1}\ge a^3-\frac{a^3b}{2\sqrt{b}}=a^3-\frac{a^3\sqrt{b}}{2}\)
Tương tự cho ta cũng có:\(\frac{b^3}{a+1}\ge b^3-\frac{b^3\sqrt{a}}{2}\)
\(\Rightarrow Q\ge a^3+b^3-\frac{a^3\sqrt{b}+b^3\sqrt{a}}{2}\ge2-\frac{a^3\sqrt{b}+b^3\sqrt{a}}{2}\left(1\right)\)
TIếp tục xài AM-GM: \(\sqrt{b}\le\frac{b+1}{2}\Rightarrow a^3\sqrt{b}=\frac{a^3b+a^3}{2}\)
\(\Rightarrow\frac{a^3\sqrt{b}+b^3\sqrt{a}}{2}\le\frac{\frac{a^3b+a^3}{2}+\frac{ab^3+b^3}{2}}{2}=\frac{a^3b+ab^3+a^3+b^3}{4}\)
\(\Rightarrow2-\frac{a^3\sqrt{b}+b^3\sqrt{a}}{2}\ge2-\frac{a^3b+ab^3+a^3+b^3}{4}\)
Cần chứng minh \(2-\frac{a^3b+ab^3+a^3+b^3}{4}\ge1\)\(\Leftrightarrow\frac{a^3b+ab^3+a^3+b^3}{4}\ge1\)
\(\Leftrightarrow a^3b+ab^3+a^3+b^3\ge4\Leftrightarrow a^3b+ab^3\ge2\) vì \(a^3+b^3\ge2\)
\(\Leftrightarrow\left(ab\right)^2\left(a+b\right)\ge2\) đúng vì ab=1 và \(a+b\ge2\)
\(\Rightarrow Q_{Min}=2-\frac{a^3\sqrt{b}+b^3\sqrt{a}}{2}\ge2-1=1\)
Khi a=b=1
