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Ta có: a + b = 2 suy ra b = a - 2
\(\Rightarrow P=\left(a^2+1\right)\left[\left(2-a\right)^2+1\right]\)
\(=\left(a^2+1\right)\left[a^2-4a+4+1\right]\)
\(=\left(a^2+1\right)\left[a^2-4a+5\right]\)
\(\Rightarrow P-5=\)\(\left(a^2+1\right)\left[a^2-4a+5\right]-5\)
\(=a^4-4a^3+6a^2-4a\)
\(=a\left(a-2\right)\left(a^2-2a+2\right)\)
\(=-a\left(2-a\right)\left(a^2-2a+1+1\right)\)
\(=-ab\left[\left(b-1\right)^2+1\right]\)
Vì a,b \(\ge\)0 và \(\left[\left(b-1\right)^2+1\right]>0\)nên
\(-ab\left[\left(b-1\right)^2+1\right]\le0\)
Vậy \(P-5\le0\Rightarrow P\le5\)
Dấu "=" khi b = 2;a = 0 và các hoán vị
