nguoivn wrote: Given $a, b, c > 0$. Prove that: $(1 + a + b + c)(1 + ab + bc + ca) \ge 4\sqrt {2}\sqrt {(a + bc)(b + ca)(c + ab)}$ Note that $(1 + a + b + c)(1 + ab + bc + ca) = (1 + a)(1 + b)(1 + c) + (a + b)(b + c)(c + a) \ge 2\sqrt {(1 + a)(1 + b)(1 + c)(a + b)(b + c)(c + a)}.$ Therefore, it suffices to prove that $8(a + bc)(b + ca)(c + ab) \le (1 + a)(1 + b)(1 + c)(a + b)(b + c)(c + a).$ By the AM-GM Inequality, we have $2\sqrt {(a + bc)(b + ca)} \le a + bc + b + ca = (c+ 1)(a + b),$ $2\sqrt {(b + ca)(c + ab)} \le (a + 1)(b + c),$ $2\sqrt {(c + ab)(a + bc)} \le (b + 1)(c + a).$ Multiplying these inequalities, we get the result.

Yes, that's also the way I created it

Let $a, b, c > 0$. Prove that $$(a+b+c)(1+ab+bc+ca)\geq \sqrt{6(a+b)(b+c)(c+a)}$$