Given: $a, b, c > 0$ and $a+b+c = 1$, prove that: $$\left(a+\dfrac{1}{b}\right)\left(b+\dfrac{1}{c}\right)\left(c+\dfrac{1}{a}\right) \ge \left(\dfrac{10}{3}\right)^3$$
Note: this is not my own post,but instead of another user that posted here at MSE but he deleted it shortly after. I think it is possibly a quite popular inequality posted in another site ( AoP ? ). In any case, I played with it and came up with an answer that I want to post here to share...
By Holder's inequality: $\sqrt[3]{\left(a+\dfrac{1}{b}\right)\left(b+\dfrac{1}{c}\right)\left(c+\dfrac{1}{a}\right)}\ge \sqrt[3]{abc}+\sqrt[3]{\dfrac{1}{abc}}=x+\dfrac{1}{x}$, where $x = \sqrt[3]{abc}\le \dfrac{a+b+c}{3} = \dfrac{1}{3}$. So $x+\dfrac{1}{x} \ge \dfrac{1}{3}+\dfrac{1}{\frac{1}{3}}= \dfrac{1}{3}+3=\dfrac{10}{3}\implies \left(a+\dfrac{1}{b}\right)\left(b+\dfrac{1}{c}\right)\left(c+\dfrac{1}{a}\right)\ge \left(\dfrac{10}{3}\right)^3$, with equality occurs when $a=b=c=\dfrac{1}{3}$.