Ta có: \(a^{3}+b^{3}=2\Leftrightarrow (a+b)(a^{2}-ab+b^{2})=2\Rightarrow a+b=\frac{2}{a^{2}-ab+b^{2}}\)
Lại có:
\(2(a-b)^{2}\geq 0\Leftrightarrow 2a^{2}-4ab+2b^{2}\geq 0\)
\(\Leftrightarrow 4a^{2}-4ab+4b^{2}\geq 2a^{2}+2b^{2}\)
\(\Leftrightarrow 4(a^{2}-ab+b^{2})\geq 2(a^{2}+b^{2})\geq (a+b)^{2}\)
\(\Leftrightarrow a^{2}-ab+b^{2}\geq \frac{(a+b)^{2}}{4}\)
\(\Rightarrow \frac{2}{a^{2}-ab+b^{2}}\leq \frac{8}{(a+b)^{2}}\Rightarrow a+b\leq \frac{8}{(a+b)^{2}}\Leftrightarrow (a+b)^{3}\leq 8\)
\(\Leftrightarrow a+b\leq 2\)
Vậy \(MAX_A=2\) \(\Leftrightarrow a=b=1\)
Đặt a = 1 + x \(\Rightarrow\) \(b^3=2-a^3\Leftrightarrow2-\left(1+x\right)^3\Leftrightarrow2-1-3x-3x^2+x^3\Leftrightarrow1-3x-3x^2+x^3\le1-3x+3x^2+x^3=\left(1-x\right)^x\)\(\Rightarrow b\le1-x\) ta có: a = 1+ x nên \(a+b\le1+x+1-x=2\)
Với a = 1 , b = 1 thì \(a^3+b^3=2\) và a + b = 2
Vậy Max A = 2 khi a=b=1