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\(4=a+b+2ab\ge2\sqrt{ab}+ab\Rightarrow ab+\sqrt{ab}-2\le0\)
\(\Rightarrow\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+2\right)\le0\)
\(\Rightarrow\sqrt{ab}-1\le0\)
\(\Rightarrow ab\le1\)
Lại có a;b ko âm nên \(ab\ge0\Rightarrow0\le ab\le1\)
\(P=a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)=\left(4-2ab\right)^3-3ab\left(4-2ab\right)\)
Đặt \(ab=x\Rightarrow0\le x\le1\)
\(P=\left(4-2x\right)^3-3x\left(4-2x\right)=-8x^3+54x^2-108x+64\)
\(=64-\frac{x}{8}\left(64x^2-432x+864\right)=64-\frac{x}{8}.\left\lbrack\left(8x-27\right)^2+135\right\rbrack\)
Do \(\left(8x-27\right)^2+135>0;\forall x\Rightarrow\frac{x}{8}\left\lbrack\left(8x-27\right)^2+135\right\rbrack\ge0;\forall x\ge0\)
\(\Rightarrow P\le64\)
\(P_{max}=64\) khi x=0 \(\Rightarrow\left(a;b\right)=\left(0;4\right);\left(4;0\right)\)
Lại có:
\(P=2+\left(-8x^3+54x^2-108x+62\right)=2+2\left(1-x\right)\left(4x^2-23x+31\right)\)
Do \(1-x\ge0;\forall x\le1\)
Đồng thời \(4x^2-23x+31=4x^2+23\left(1-x\right)+8>0;\forall x\le1\)
\(\Rightarrow2\left(1-x\right)\left(4x^2-23x+31\right)\ge0;\forall x\le1\)
\(\Rightarrow P\ge2\)
\(P_{\min}=2\) khi x=1 =>a=b=1
đặt \(t=a+b\) từ GT => \(3=t^2-ab\ge\frac{3}{4}t^2\)\(\Leftrightarrow\)\(-2\le t\le2\)
\(P=-4t^3-3t^2+18t+9=\hept{\begin{cases}\frac{-1}{4}\left(2t+3\right)^2\left(4t-9\right)-\frac{45}{4}\ge\frac{-45}{4}\left(dungvoit\le2\right)\\-\left(t-1\right)^2\left(4t+11\right)+20\le20\left(dungvoit\ge-2\right)\end{cases}}\)
\(P_{min}=\frac{-45}{4}\) tại
\(\hept{\begin{cases}a^2+b^2+ab=3\\a+b=\frac{-3}{2}\end{cases}}\Leftrightarrow\left(a;b\right)=\left\{\left(\frac{-3-\sqrt{21}}{4};\frac{-3+\sqrt{21}}{4}\right);\left(\frac{-3+\sqrt{21}}{4};\frac{-3-\sqrt{21}}{4}\right)\right\}\)
\(P_{max}=20\) tại \(\hept{\begin{cases}a^2+b^2+ab=3\\a+b=1\end{cases}}\Leftrightarrow\left(a;b\right)=\left\{\left(2;-1\right);\left(-1;2\right)\right\}\)
Có: \(4=\left(a+b\right)^2-\left(b-1\right)^2\le\left(a+b\right)^2\)\(\Rightarrow\)\(a+b\ge2\)
\(P=\frac{\frac{a^4}{a}+\frac{b^4}{b}}{ab}\ge\frac{\frac{\left(a^2+b^2\right)^2}{a+b}}{ab}\ge\frac{\frac{\left[\frac{\left(a+b\right)^2}{2}\right]^2}{a+b}}{ab}=\frac{\left(a+b\right)\left(a+b\right)^2}{4ab}\ge\frac{2\left(2\sqrt{ab}\right)^2}{4ab}=2\)
"=" \(\Leftrightarrow\)\(a=b=1\)