\(2\left(a^2+b^2\right)\ge\left(a+b\right)^2\Leftrightarrow4\ge\left(a+b\right)^2\Leftrightarrow-2\le a+b\le2\)

\(4ab\le2\left(a^2+b^2\right)\Leftrightarrow4ab\le4\Leftrightarrow ab\le1\)

\(A=\left(3-a\right)\left(3-b\right)=9-3a-3b+ab=9-3\left(a+b\right)+ab\le9+3.2+1=16\)

\(A_{max}=16\Leftrightarrow a=b=-1\)

\(\left(a+b\right)^2\ge2\left(a^2+b^2\right)=4\Rightarrow-2\le a+b\le2\)

\(P=9-3\left(a+b\right)+ab=9-3\left(a+b\right)+\dfrac{\left(a+b\right)^2-\left(a^2+b^2\right)}{2}\)

\(P=\dfrac{1}{2}\left(a+b\right)^2-3\left(a+b\right)+8\)

Đặt \(a+b=x\Rightarrow-2\le x\le2\)

\(P=\dfrac{1}{2}x^2-3x+8=\dfrac{1}{2}\left(x-2\right)\left(x-4\right)+4\)

Do \(-2\le x\le2\Rightarrow\left\{{}\begin{matrix}x-2\le0\\x-4< 0\end{matrix}\right.\) \(\Rightarrow\left(x-2\right)\left(x-4\right)\ge0\)

\(\Rightarrow P\ge4\Rightarrow P_{min}=4\) khi \(x=2\Leftrightarrow a=b=1\)

\(P=\dfrac{1}{2}\left(x+2\right)\left(x-8\right)+16\)

Do \(-2\le x\le2\Rightarrow\left\{{}\begin{matrix}x+2\ge0\\x-8< 0\end{matrix}\right.\) \(\Rightarrow\dfrac{1}{2}\left(x+2\right)\left(x-8\right)\le0\)

\(\Rightarrow P\le16\Rightarrow P_{max}=16\) khi \(x=-2\Leftrightarrow a=b=-1\)

\(A=a^6+b^6=\left(a^2\right)^3+\left(b^2\right)^3\)

\(=\left(a^2+b^2\right)\left(a^4+b^4-a^2b^2\right)\)

\(=1.\left[\left(a^4+b^4+2a^2b^2\right)-3a^2b^2\right]\)

\(=\left(a^2+b^2\right)^2-3a^2b^2\)

\(=1^2-3a^2b^2\)

\(\left(a-b\right)^2\ge0\Rightarrow a^2+b^2-2ab\ge0\Rightarrow\frac{a^2+b^2}{2}\ge ab\)

\(\Rightarrow ab\le1:2=0,5\Rightarrow3a^2b^2\le\frac{3}{4}\)

\(\Rightarrow A=1^2-3a^2b^2\ge1-\frac{3}{4}=\frac{1}{4}\)

\(\Rightarrow MinA=\frac{1}{4}\Leftrightarrow a=b=\frac{1}{2}\)

Vậy ...

\(\left(a^3+b^3\right)\left(a+b\right)=ab\left(1-a\right)\left(1-b\right)\)

\(\Leftrightarrow\left(1-a\right)\left(1-b\right)=\left(\dfrac{a^2}{b}+\dfrac{b^2}{a}\right)\left(a+b\right)\ge\left(a+b\right)^2\ge4ab\)

\(\Rightarrow1+ab-4ab\ge a+b\ge2\sqrt{ab}\)

\(\Rightarrow3ab+2\sqrt{ab}-1\le0\)

\(\Leftrightarrow\left(\sqrt{ab}+1\right)\left(3\sqrt{ab}-1\right)\le0\)

\(\Leftrightarrow ab\le\dfrac{1}{9}\)