\(\dfrac{a+1}{1+b^2}=a+1-\dfrac{\left(a+1\right)b^2}{1+b^2}\ge a+1-\dfrac{\left(a+1\right)b^2}{2b}=a+1-\dfrac{1}{2}\left(ab+b\right)\)
Tương tự, cộng lại:
\(P\ge a+b+c+3-\dfrac{1}{2}\left(ab+bc+ca+a+b+c\right)=6-\dfrac{1}{2}\left(ab+bc+ca+3\right)\)
\(P\ge\dfrac{9}{2}-\dfrac{1}{2}\left(ab+bc+ca\right)\ge\dfrac{9}{2}-\dfrac{1}{6}\left(a+b+c\right)^2=3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
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