Let $\sqrt[3]{\frac ab}=x$ and $\sqrt[3]{\frac ba}=y$. Note that $xy=1$. After cubing both sides it's equivalent to prove $(x+y)^3+4 \ge 6(x+y)$. AM-GM $\implies \frac{(x+y)^3}{4}+2+2\ge 3(x+y)$ AM-GM $\implies x+y\ge 2\sqrt{xy}=2 \implies \frac 34 (x+y)^3 \ge 3(x+y).$ So we are done.

Let $a,b$ and $c$ be positive numbers. Prove the inequality $$\sqrt[3]{\frac ab}+\sqrt[3]{\frac bc}+\sqrt[3]{\frac ca}\le\sqrt[3]{3(a+b+c)\left(\frac1a+\frac1b+\frac1c\right)}.$$Let $a_1,a_2,\cdots,a_n (n\ge 2)$ be positive numbers. Prove that$$\sqrt[3]{\frac {a_1}{a_2}}+\sqrt[3]{\frac {a_2}{a_3}}+\cdots+\sqrt[3]{\frac {a_{n-1}}{a_n}}+\sqrt[3]{\frac {a_n}{a_1}}\leq \sqrt[3]{n(a_1+a_2+\cdots+a_n)\left(\frac1{a_1}+\frac1{a_2}+\cdots+\frac1{a_n}\right)}.$$

Could someone tell me where did the inequality $\frac 34 (x+y)^3 \ge 3(x+y)$ follows from? I'm getting by AM-GM $\frac{3(x+y)^{3}}{4}+12 \ge 9(x+y)$...

It's because \(\frac14 (x+y)^2 \ge xy = 1.\)

sqing wrote: Let $a_1,a_2,\cdots,a_n (n\ge 2)$ be positive numbers. Prove that$$\sqrt[3]{\frac {a_1}{a_2}}+\sqrt[3]{\frac {a_2}{a_3}}+\cdots+\sqrt[3]{\frac {a_{n-1}}{a_n}}+\sqrt[3]{\frac {a_n}{a_1}}\leq \sqrt[3]{n(a_1+a_2+\cdots+a_n)\left(\frac1{a_1}+\frac1{a_2}+\cdots+\frac1{a_n}\right)}.$$