Ta co:
\(\left(ab+bc+ca\right)^2\left(a^2+b^2+c^2\right)\)
\(=\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)\left(ab+bc+ca\right)\le\text{ }\frac{\left[a^2+b^2+c^2+2\left(ab+bc+ca\right)\right]^3}{27}\)
\(\frac{\left[a^2+b^2+c^2+2\left(ab+bc+ca\right)\right]^3}{27}=\frac{\left(a+b+c\right)^6}{27}=\frac{3^6}{27}=27\)
Dau '=' xay ra khi \(a=b=c=1\)