Câu 3:
Cách 2: Theo BĐT Cô-si (Cauchy, AM-GM) ta có
$$\dfrac{a^3}{b}+ab+\dfrac{b^3}{c}+bc+\dfrac{c^3}{a}+ac\ge 2(a^2+b^2+c^2)$$
Cũng có
$$\dfrac{a^3}{b}+bc+\dfrac{b^3}{c}+ac+\dfrac{c^3}{a}+ab\ge 2\left( a\sqrt{ac}+b\sqrt{ab}+c\sqrt{bc}\right)$$
Cộng vế suy ra
$$\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}+ab+bc+ac\ge a\sqrt{ac}+b\sqrt{ab}+c\sqrt{bc}+a^2+b^2+c^2$$
$$\Leftrightarrow \dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge a\sqrt{ac}+b\sqrt{ab}+c\sqrt{bc}+(a^2+b^2+c^2-ab-ac-bc)\ge a\sqrt{ac}+b\sqrt{ab}+c\sqrt{bc}.$$