It's equivalent to $\frac{a(c+1)+b(a+1)+c(b+1)}{(a+1)(b+1)(c+1)}\ge \frac{3}{4}\Leftrightarrow 4a(c+1)+4b(a+1)+4c(b+1)\ge 3(a+1)(b+1)(c+1)$ $\Leftrightarrow ab+bc+ca+a+b+c\ge 6$ (since $abc=1$) By AM-GM, $$ab+bc+ca+a+b+c\ge 3\sqrt[3]{a^2b^2c^2}+3\sqrt[3]{abc}=6$$ Done

parmenides51 wrote: Prove that $\frac{a}{(a + 1)(b + 1)} +\frac{ b}{(b + 1)(c + 1)} + \frac{c}{(c + 1)(a + 1)} \ge \frac34$ where $a, b$ and $c$ are positive real numbers satisfying $abc = 1$. https://artofproblemsolving.com/community/c6h326940p1751962 https://artofproblemsolving.com/community/c3h1092498p4868541

BestChoice123 wrote: It's equivalent to $\frac{a(c+1)+b(a+1)+c(b+1)}{(a+1)(b+1)(c+1)}\ge \frac{3}{4}\Leftrightarrow 4a(c+1)+4b(a+1)+4c(b+1)\ge 3(a+1)(b+1)(c+1)$ $\Leftrightarrow ab+bc+ca+a+b+c\ge 3$ (since $abc=1$) By AM-GM, $$ab+bc+ca+a+b+c\ge 3\sqrt[3]{a^2b^2c^2}+3\sqrt[3]{abc}=6$$ Done are you typo $\Leftrightarrow ab+bc+ca+a+b+c\ge 3$ (since $abc=1$)? Maybe $\Leftrightarrow ab+bc+ca+a+b+c\ge 6$ (since $abc=1$)

Mathskidd wrote: BestChoice123 wrote: It's equivalent to $\frac{a(c+1)+b(a+1)+c(b+1)}{(a+1)(b+1)(c+1)}\ge \frac{3}{4}\Leftrightarrow 4a(c+1)+4b(a+1)+4c(b+1)\ge 3(a+1)(b+1)(c+1)$ $\Leftrightarrow ab+bc+ca+a+b+c\ge 3$ (since $abc=1$) By AM-GM, $$ab+bc+ca+a+b+c\ge 3\sqrt[3]{a^2b^2c^2}+3\sqrt[3]{abc}=6$$ Done are you typo $\Leftrightarrow ab+bc+ca+a+b+c\ge 3$ (since $abc=1$)? Maybe $\Leftrightarrow ab+bc+ca+a+b+c\ge 6$ (since $abc=1$) Oh I see it Thx