Áp dụng bđt AM - GM, ta có:
\(\circledast\) \(abc\le\dfrac{\left(a+b+c\right)^3}{27}=\dfrac{1}{27}\)
\(\circledast\) \(\left(a+b+c\right)^3-a^3-b^3-c^3=3\left(b+c\right)\left(c+a\right)\left(a+b\right)\)
\(\Rightarrow a^3+b^3+c^3=1-3\left(b+c\right)\left(c+a\right)\left(a+b\right)\)
\(\le1-3\times8abc=1-24abc\)
Áp dụng bđt Cauchy Shwarz dạng Engel, ta có:
\(\dfrac{a}{a^2+1}+\dfrac{b}{b^2+1}+\dfrac{c}{c^2+1}+\dfrac{1}{9abc}\)
\(=\dfrac{a^2}{a^3+a}+\dfrac{b^2}{b^3+b}+\dfrac{c^2}{c^3+c}+\dfrac{1}{9abc}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{a^3+b^3+c^3+1}+\dfrac{1}{9abc}\)
\(\ge\dfrac{1}{2-24abc}+\dfrac{1}{30abc}+\dfrac{7}{90abc}\)
\(\ge\dfrac{\left(1+1\right)^2}{6abc+2}+\dfrac{7}{90abc}\ge\dfrac{4}{6\times\dfrac{1}{27}+2}+\dfrac{7}{90\times\dfrac{1}{27}}=\dfrac{39}{10}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)