Write it like this $(\frac{1}{4a}+\frac{1}{4a}+\frac{1}{4b}+\frac{1}{4b}+\frac{1}{4c}+\frac{1}{4c}+\frac{1}{2abc})(ab^2+ac^2+ba^2+bc^2+ca^2+cb^2+2abc)\ge(a+b+c+1)^2$ Then Cauchy-Inequality easy kills problem. Pretty similar problem: $2(a^2+1)(b^2+1)(c^2+1)\ge(a+1)(b+1)(c+1)(a+b+c-1)$ For all $a,b,c\ge0$

JosefSvejk wrote: Prove for positive reals $a,b,c$ that $(ab+bc+ca+1)(a+b)(b+c)(c+a) \ge 2abc(a+b+c+1)^2$ It came from the Potla's known inequality: $$\frac{ab+bc+ca+1}{(a+b+c+1)^2}+\frac 38\sqrt[3]{\frac{(a+b)(b+c)(c+a)}{abc}}\geq 1.$$

arqady wrote: JosefSvejk wrote: Prove for positive reals $a,b,c$ that $(ab+bc+ca+1)(a+b)(b+c)(c+a) \ge 2abc(a+b+c+1)^2$ It came from the Potla's known inequality: $$\frac{ab+bc+ca+1}{(a+b+c+1)^2}+\frac 38\sqrt[3]{\frac{(a+b)(b+c)(c+a)}{abc}}\geq 1.$$ looks like a difficult task