Đặt \(A=a^3+b^3+c^3-3\left(a-1\right)\left(b-1\right)\left(c-1\right)\)
\(=\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)-3\left(a-1\right)\left(b-1\right)\left(c-1\right)\)
\(=3^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)-3\left(a-1\right)\left(b-1\right)\left(c-1\right)\)
\(=27-3\left(a+b\right)\left(b+c\right)\left(c+a\right)-3\left(a-1\right)\left(b-1\right)\left(c-1\right)\)
Áp dụng BĐT Caushy ta có:
\(A=27-3\left(a+b\right)\left(b+c\right)\left(c+a\right)-3\left(a-1\right)\left(b-1\right)\left(c-1\right)\ge27-3.\left[\dfrac{\left(a+b\right)+\left(b+c\right)+\left(c+a\right)}{3}\right]^3-3\left[\dfrac{\left(a-1\right)+\left(b-1\right)+\left(c-1\right)}{3}\right]^3=27-3.\left[\dfrac{2\left(a+b+c\right)}{3}\right]^3-3\left[\dfrac{\left(a+b+c\right)-3}{3}\right]^3=27-3.\left(\dfrac{2.3}{3}\right)^3-3\left(\dfrac{3-3}{3}\right)^3=3\)\(\Rightarrow A\ge3\left(1\right)\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\).
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Theo đề ta có: \(\left(2-a\right)\left(2-b\right)\left(2-c\right)\ge0\)
\(\Rightarrow-4\left(a+b+c\right)+2\left(ab+bc+ca\right)-abc+8\ge0\)
\(\Rightarrow-4.3+2\left(ab+bc+ca\right)-abc+8\ge0\)
\(\Rightarrow2\left(ab+bc+ca\right)-abc-4\ge0\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)-abc-4\ge a^2+b^2+c^2\)
\(\Rightarrow a^2+b^2+c^2\le\left(a+b+c\right)^2-4-abc=3^3-4-abc=5-abc\)
\(\Rightarrow a^2+b^2+c^2\le5\) (do \(abc\ge0\))
\(A=a^3+b^3+c^3-3\left(a-1\right)\left(b-1\right)\left(c-1\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)-3\left[\left(a+b+c\right)-\left(ab+bc+ca\right)+abc-1\right]\)
\(=3.\left(a^2+b^2+c^2-ab-bc-ca\right)-3\left[3-\left(ab+bc+ca\right)+abc-1\right]\)
\(=3\left(a^2+b^2+c^2\right)-3\left(ab+bc+ca\right)+3\left(ab+bc+ca\right)-3abc-6\)
\(=3\left(a^2+b^2+c^2\right)-3abc-6\le3.5-3.0-6=9\)
\(\Rightarrow A\le9\left(2\right)\)
Dấu "=" xảy ra \(\Leftrightarrow\left(a,b,c\right)=\left(0,1,2\right)\) và các hoán vị.
Từ (1), (2) suy ra: \(3\le A\le9\Rightarrowđpcm\)