Ta có: \(\frac{a^3}{b}+ab\ge2\cdot\sqrt{\frac{a^3}{b}\cdot ab}=2\sqrt{a^4}=2a^2\)
\(\frac{b^3}{c}+bc\ge2\cdot\sqrt{\frac{b^3}{c}\cdot bc}=2\sqrt{b^4}=2b^2\)
\(\frac{c^3}{a}+ca\ge2\cdot\sqrt{\frac{c^3}{a}\cdot ca}=2\cdot\sqrt{c^4}=2c^2\)
Do đó: \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}+ab+bc+ac\ge2\left(a^2+b^2+c^2\right)\left(1\right)\)
Ta có: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\forall a,b,c>0\)
=>\(a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ac+c^2\ge0\)
=>\(2a^2+2b^2+2c^2\ge2ab+2ac+2bc=2\left(ab+ac+bc\right)\) (2)
Từ (1),(2) suy ra \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}+ab+ac+bc\ge2\cdot\left(ab+ac+bc\right)\)
=>\(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge ab+bc+ac\)
