cho a,b,c tm a+b+c >0 và -1 < a,b,c < 1
cm \(^{a^2}\)+\(^{b^2}\)+\(^{c^2}\) <2
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\(4b.ac+\left(a+c\right)^2\le4b.\dfrac{1}{4}\left(a+c\right)^2+\left(a+c\right)^2=\left(a+c\right)^2\left(b+1\right)\)
\(\Rightarrow T\ge\dfrac{1}{\left(a+c\right)^2}+\dfrac{1}{\left(a+b\right)^2}\ge\dfrac{1}{2\left(a^2+c^2\right)}+\dfrac{1}{2\left(a^2+b^2\right)}\ge\dfrac{4}{2\left(2a^2+b^2+c^2\right)}\)
\(\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2}\)
\(=\dfrac{a^3+ab^2-ab^2}{a^2+b^2}+\dfrac{b^3+bc^2-bc^2}{b^2+c^2}+\dfrac{c^3+ca^2-ca^2}{c^2+a^2}\)
\(=\dfrac{a\left(a^2+b^2\right)}{a^2+b^2}+\dfrac{b\left(b^2+c^2\right)}{b^2+c^2}+\dfrac{c\left(c^2+a^2\right)}{c^2+a^2}-\left(\dfrac{ab^2}{a^2+b^2}+\dfrac{bc^2}{b^2+c^2}+\dfrac{ca^2}{c^2+a^2}\right)\)\(=a+b+c-\left(\dfrac{ab^2}{a^2+b^2}+\dfrac{bc^2}{b^2+c^2}+\dfrac{ca^2}{c^2+a^2}\right)\)\(\ge6-\left(\dfrac{ab^2}{2ab}+\dfrac{bc^2}{2bc}+\dfrac{ca^2}{2ca}\right)=6-\dfrac{a+b+c}{2}=6-\dfrac{6}{2}=3\) (dpcm)Dau = xay ra khi a=b=c=2
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