sharifymatholympiad wrote: Positive real numbers $a,b,c$ are given such that $abc=1$. Prove that$$a+b+c+\frac{3}{ab+bc+ca}\geq4.$$ https://artofproblemsolving.com/community/c6h623571p4627645 Positive real numbers $a,b,c$ are given such that $abc=1$. Prove that$$\frac{a+1}{b(c^2+1)}+\frac{b+1}{c(a^2+1)}+\frac{c+1}{a(b^2+1)}\ge 3$$

By homogenizing the inequality, it's equivalent to showing \[(a+b+c)(ab+bc+ca)+3abc\geqslant 4\sqrt[3]{abc}(ab+bc+ca)\]or \[\sum_{cyc}a^2b+3abc+\sum_{cyc} ab^2+3abc\geqslant 4\sum_{cyc} a^{\frac{4}{3}}b^{\frac{4}{3}}c\]which follows from adding up two Schur's 3rd degree inequalities $(a^{\frac{2}{3}}b^{\frac{1}{3}},b^{\frac{2}{3}}c^{\frac{1}{3}},c^{\frac{2}{3}}a^{\frac{1}{3}}),(b^{\frac{2}{3}}a^{\frac{1}{3}},c^{\frac{2}{3}}b^{\frac{1}{3}},a^{\frac{2}{3}}c^{\frac{1}{3}}):$ \[\sum_{cyc}a^2b+3abc\geqslant 2\sum_{cyc} a^{\frac{4}{3}}b^{\frac{4}{3}}c\]\[\sum_{cyc} ab^2+3abc\geqslant 2\sum_{cyc} a^{\frac{4}{3}}b^{\frac{4}{3}}c.\]

sharifymatholympiad wrote: Positive real numbers $a,b,c$ are given such that $abc=1$. Prove that $$a+b+c+\frac{3}{ab+bc+ca}\geq4.$$ Let $a+b+c=x$. By AM-GM $x\ge 3$. We have $a+b+c+\frac{9}{3(ab+bc+ca)}\ge\frac{x}{3}+\left(\frac{x}{3}+\frac{x}{3}+\frac{9}{x^2}\right)\ge 1+3$. This approach works also for $a+b+c+\frac{9}{2(ab+bc+ca)}\ge\frac{9}{2}$.

Excellent