This problem was proposed by me. I intended to send it to the Baltic Way this year, but later I sadly discovered that Balkan Shortlist 2018 A4 uses the main claims, therefore it was proposed to TST instead.

Another sol. We want \[ 2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\ge a+b+c+\frac1a+\frac1b+\frac1c, \]where $1/a+1/b+1/c = ab+bc+ca$ as $abc=1$. Note that $a/b + b/c + c/a = ab^2+bc^2+ca^2$. Using Cauchy-Schwarz twice \[ (ab^2+bc^2+ca^2)(a+b+c)\ge (ab+bc+ca)^2 \quad\text{and}\quad (ab^2+bc^2+ca^2)\left(\frac1a+\frac1b+\frac1c\right)\ge (a+b+c)^2. \]Setting $m=ab+bc+ca$ and $n=a+b+c$, this implies \[ 2(ab^2+bc^2+ca^2) \ge \frac{m^2}{n}+\frac{n^2}{m} \ge m+n = a+b+c+\frac1a+\frac1b+\frac1c, \]where we used $m^2/n+n^2/m\ge m+n$.

Let $ a, b, c$ be positive real numbers such that $ abc = 1. $ Prove that: $$ 2 (a^ 2 + b^ 2 + c^ 2) \left (\frac 1 {a^ 2} + \frac 1{b^ 2}+ \frac 1{c^2}\right)\geq 3(a+ b + c +\frac1a+\frac1b+\frac1c)$$ Kimchiks926 wrote: Prove that for positive real numbers $a,b,c$ satisfying $abc=1$ the following inequality holds: $$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a} \ge \frac{a^2+1}{2a}+\frac{b^2+1}{2b}+\frac{c^2+1}{2c}.$$ $$\iff$$$$2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\ge a+b+c+\frac1a+\frac1b+\frac1c$$