Ta có: \(\dfrac{b}{a^2+1}=b-\dfrac{a^2b}{a^2+1}\ge b-\dfrac{a^2b}{2a}=b-\dfrac{ab}{2}\)
Tương tự và suy ra: \(P=\dfrac{b}{a^2+1}+\dfrac{c}{b^2+1}+\dfrac{a}{c^2+1}+\dfrac{1}{4}\left(ab+bc+ca\right)\ge a+b+c-\dfrac{1}{2}\left(ab+bc+ca\right)+\dfrac{1}{4}\left(ab+bc+ca\right)=a+b+c-\dfrac{1}{4}\left(ab+bc+ca\right)\)
Để ý \(ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}\), do đó:
\(P\ge a+b+c-\dfrac{1}{4}.\dfrac{\left(a+b+c\right)^2}{3}=3-\dfrac{1}{4}.\dfrac{9}{3}=\dfrac{9}{4}\)
Vậy MinP=9/4 khi a=b=c=1.
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