Let $a_{1},a_{2},\cdots,a_{n}>0(n>1)$,such $a_{1}a_{2}\cdots a_{n}=1$,show that $$\dfrac{a_{1}}{a_{2}}+\dfrac{a_{2}}{a_{3}}+\cdots+\dfrac{a_{n}}{a_{1}}\ge a_{1}+a_{2}+\cdots+a_{n}$$
when $n=2$,$$\dfrac{a_{1}}{a_{2}}+\dfrac{a_{2}}{a_{1}}\ge a_{1}+a_{2}$$ or $$a^2_{1}+a^2_{2}\ge\dfrac{1}{2}(a_{1}+a_{2})^2\ge a_{1}+a_{2}$$ since $$a_{1}+a_{2}\ge 2\sqrt{a_{1}a_{2}}=2$$ and $n=3$,because use AM-GM $$\dfrac{a_{1}}{a_{2}}+\dfrac{a_{1}}{a_{2}}+\dfrac{a_{2}}{a_{3}}\ge 3\sqrt[3]{\dfrac{a^2_{1}}{a_{2}a_{3}}}=3a_{1}$$ so $$\sum_{cyc}\dfrac{a_{1}}{a_{2}}+\dfrac{a_{1}}{a_{2}}+\dfrac{a_{2}}{a_{3}}\ge 3\sum_{cyc}a_{1}$$ so $$\dfrac{a_{1}}{a_{2}}+\dfrac{a_{2}}{a_{3}}+\dfrac{a_{3}}{a_{1}}\ge a_{1}+a_{2}+a_{3}$$ But for $n\ge 4$,I can't prove it.Thanks
If n=2m and $a_1=a_2=...=a_m=2,a_{m+1}=..=a_{2m}=\frac12$. The left side is $2(m-1)+4+\frac14$ but the right side is $\frac52 m$
– Zhaohui Du Aug 24 '19 at 09:36