Đề giả thiết cho như vậy hay là \(a^3+b^3+6ab\le8???\)
Ta có \(a^3+b^3+8ab\le10\)
Áp dụng cosi ta có \(a^3+b^3+1\ge3ab\)
=> \(11ab\le11\)=> \(ab\le1\)
+ \(a^3+a^3+1\ge3a^2\); \(b^3+b^3+1\ge3b^2\)
=> \(2a^3+2b^3+2\ge3\left(a^2+b^2\right)\)
=> \(a^3+b^3\ge\frac{3\left(a^2+b^2\right)-2}{2}\)
=> \(3\left(a^2+b^2\right)+16ab\le22\)
=> \(P\ge\frac{3}{22-16ab}+\frac{5}{ab}+3ab=\left(\frac{3}{22-16ab}+\frac{22-16ab}{12}\right)+5\left(\frac{1}{ab}+ab\right)-\frac{22}{12}-\frac{2}{3}ab\)
=> \(P\ge2\sqrt{\frac{3}{12}}+5.2-\frac{22}{12}-\frac{2}{3}.1\)
=> \(P\ge\frac{17}{2}\)
Vậy MinP=17/2 khi a=b=1
