\(\Leftrightarrow1+b^2+a^2\left(b^3+b\right)\le\left(2b^3+2\right)a^2-2\left(b^3+1\right)a+2b^3+2\)

\(\Leftrightarrow\left(b^3-b+2\right)a^2-2\left(b^3+1\right)a+2b^3-b^2+1\ge0\)

Xét tam thức bậc 2: \(f\left(a\right)=\left(b^3-b+2\right)a^2-2\left(b^3+1\right)a+2b^3-b^2+1\)

Ta có: \(b^3+2-b\ge3b-b=2b>0\)

\(\Delta'=\left(b^3+1\right)^2-\left(b^3-b+2\right)\left(2b^3-b^2+1\right)\)

\(\Delta'=-\left(b-1\right)^2\left(b^4+b^3-b^2+b+1\right)\le0\) ; \(\forall b>0\)

\(\Rightarrow f\left(a\right)\ge0\) ; \(\forall a\)

Dấu "=" xảy ra khi \(\left(a;b\right)=\left(1;1\right)\)

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

Vì \(a^2-ab+b^2=\left(a-\frac{b}{2}\right)^2+\frac{3b^2}{4}\ge0\)

Nên a + b > 0

*Vì a + b > 0

\(\Rightarrow\left(a-b\right)^2\left(a+b\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2-b^2\right)\ge0\)

\(\Leftrightarrow a^3+b^3-ab\left(a+b\right)\ge0\)

\(\Leftrightarrow a^3+b^3\ge ab\left(a+b\right)\)

\(\Leftrightarrow3\left(a^3+b^3\right)\ge3ab\left(a+b\right)\)

\(\Leftrightarrow4\left(a^3+b^3\right)\ge a^3+b^3+3ab\left(a+b\right)\)

\(\Leftrightarrow4.2\ge\left(a+b\right)^3\)

\(\Leftrightarrow2\ge a+b\)

Vậy .....

\(a^3+b^3=2\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)=2\Leftrightarrow\left(a+b\right)\left[\left(a-\frac{b}{2}\right)^2+\frac{3b^2}{4}\right]=2.\) 

Suy ra  :  a+b > 0  

Bài 2: 

  Đặt   \(a=3+x\)và   \(b=3+y\)thì    \(x,y\ge0\). Ta có :  \(a+b=6+\left(x+y\right)\).

Ta cần chứng minh   \(x+y\ge1\)

Ví dụ   \(x+y< 1\)thì  \(x^2+2xy+y^2< 1\)nên \(x^2+y^2< 1\)

\(\Leftrightarrow a^2+b^2=\left(x+3\right)^2+\left(y+3\right)^2=18+6\left(x+y\right)+\left(x^2+y^2\right)< 18+6+1=25\)

Điều này ngược với  giả thiết ở đề bài   \(ầ^2+b^2\ge25\)

Vậy \(x+y\ge1\)\(\Leftrightarrow a+b\ge7\left(dpcm\right)\)

tk mk nka !!!

Vì \(0\le a\le2;0\le b\le2;0\le c\le2\Rightarrow\left(2-a\right)\left(2-b\right)\left(2-c\right)\ge0\)\(\Leftrightarrow8-4\left(a+b+c\right)+2\left(ab+bc+ca\right)-abc\ge0\)\(\Leftrightarrow2\left(ab+bc+ca\right)\ge4\left(a+b+c\right)-8+abc\ge4\)\(\Leftrightarrow2\left(ab+bc+ca\right)\ge12-8+abc\ge4\)

\(\Rightarrow\)\(2\left(ab+bc+ca\right)\ge4\)

\(\Leftrightarrow-2\left(ab+bc+ca\right)\le-4\)

Ta có :

\(a+b+c=3\Rightarrow\left(a+b+c\right)^2=9\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=9\)

\(\Rightarrow a^2+b^2+c^2=9-2\left(ab+bc+ca\right)\le9-4=5\Rightarrowđpcm\)Đẳng thức xảy ra khi

\(\left(2-a\right)\left(2-b\right)\left(2-c\right)=0\)

\(\left[{}\begin{matrix}2-a=0\\2-b=0\\2-c=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=2\\b=2\\c=2\end{matrix}\right.\)