$\Big(ab + bc +\frac{1}{ca}\Big)\Big(bc + ca +\frac{1}{ab}\Big)\Big(ca + ab +\frac{1}{bc}\Big)= b(a+c+1)c(b+a+1)a(c+b+1)=(a+c+1)(b+a+1)(c+b+1)$ $(a+c+1)(b+a+1)(c+b+1)- (1 + 2a)(1 + 2b)(1 + 2c)=a^2b+b^2a+a^2c+c^2a+b^2c+c^2b+a^2+b^2+c^2-ab-ac-bc-6abc \geq 0$

parmenides51 wrote: Let $a, b$ and $c$ be positive real numbers such that $abc = 1$. Prove the inequality $\Big(ab + bc +\frac{1}{ca}\Big)\Big(bc + ca +\frac{1}{ab}\Big)\Big(ca + ab +\frac{1}{bc}\Big)\ge (1 + 2a)(1 + 2b)(1 + 2c)$. https://artofproblemsolving.com/community/c6h502853p2825205

Let $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$, then multiply $x^2y^2z^2$, we get $(\sum_{cyc}x^4z^2)+(\sum_{cyc}x^4yz)+(\sum_{cyc}x^3y^3)$ $\geq$ $(\sum_{cyc}x^3y^2z)+6x^2y^2z^2$ This holds by AM-GM inequality

parmenides51 wrote: Let $a, b$ and $c$ be positive real numbers such that $abc = 1$. Prove the inequality $\Big(ab + bc +\frac{1}{ca}\Big)\Big(bc + ca +\frac{1}{ab}\Big)\Big(ca + ab +\frac{1}{bc}\Big)\ge (1 + 2a)(1 + 2b)(1 + 2c)$. $\Big(ab + bc +\frac{1}{ca}\Big)\Big(bc + ca +\frac{1}{ab}\Big) \ge (2+bc)^2$ $\Big(bc + ca +\frac{1}{ab}\Big)\Big(ca + ab +\frac{1}{bc}\Big) \ge (2+ca)^2$ $\Big(ca + ab +\frac{1}{bc}\Big)\Big(ab + bc +\frac{1}{ca}\Big) \ge (2+ab)^2$ $\Big(ab + bc +\frac{1}{ca}\Big)\Big(bc + ca +\frac{1}{ab}\Big)\Big(ca + ab +\frac{1}{bc}\Big)\ge (2+ab)(2+bc)(2+ca)= (1 + 2a)(1 + 2b)(1 + 2c)$