\(\left(2-a\right)\left(2-b\right)\left(2-c\right)\ge0\)
\(\Leftrightarrow8-4\left(a+b+c\right)+2\left(ab+bc+ca\right)-abc\ge0\)
\(\Leftrightarrow2\left(ab+bc+ca\right)\ge4\left(a+b+c\right)-8+abc\)
\(\Leftrightarrow2\left(ab+bc+ca\right)\ge12-8+abc\ge4\)
\(\Rightarrow2\left(ab+bc+ca\right)\ge4\)
\(\Rightarrow-2\left(ab+bc+ca\right)\le-4\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)=9\)
\(\Rightarrow a^2+b^2+c^2=9-2\left(ab+bc+ca\right)\le9-4=5\)(Đpcm)
Dấu = khi \(\hept{\begin{cases}\left(2-a\right)\left(2-b\right)\left(2-c\right)=0\\abc=0\\a+b+c=3\end{cases}}\)
\(\Rightarrow\left(a;b;c\right)=\left(2;1;0\right)\)và hoán vị.
a = 2 ( t/m )
b = 1 ( t/m )
c = 0 ( t/m )
vậy \(a^2+b^2+c^2\le5\)
