Apply Cauchy-Schwarz: \[ \sum_{{\rm cyc}}\frac{a}{b(1+c)} = \sum_{{\rm cyc}}\frac{a^2}{ab+abc}\ge \frac{(a+b+c)^2}{ab+bc+ca+3}. \]With this, it boils down proving \[ 2(a+b+c)^2 \ge 3(ab+bc+ca)+9\iff 2(a^2+b^2+c^2)+(ab+bc+ca)\ge 9. \]Since $\min\{a^2+b^2+c^2,ab+bc+ca\}\ge 3$ trivially by AM-GM, we are done.

Kimchiks926 wrote: Prove that for positive real numbers $a,b,c$ satisfying $abc=1$ the following inequality holds: $$ \frac{a}{b(1+c)} +\frac{b}{c(1+a)}+\frac{c}{a(1+b)} \ge \frac{3}{2} $$ https://artofproblemsolving.com/community/c6h1219608p6090513

Note that $\frac{a}{b(1+c)}=\frac{a^2}{ab+1}$ and $\frac{b}{c(1+a)}=\frac{b^2}{bc+1}$ and $\frac{c}{a(1+b)}=\frac{c^2}{ac+1}$ Thus by titu's lemma: $\frac{a}{b(1+c)} +\frac{b}{c(1+a)}+\frac{c}{a(1+b)}=\frac{a^2}{ab+1}+\frac{b^2}{bc+1}+\frac{c^2}{ac+1} \ge \frac{(a+b+c)^2}{bc+ac+ab+3}$ Now note that the problem is equivalent to $2(a^2+b^2+c^2)+ab+bc+ac+3 \ge 9$ but this is true since by AM-GM: $2(a^2+b^2+c^2)+ab+bc+ac+3 \ge 9 \sqrt[3]{(abc)^2}=9$ And equality holds when $a=b=c=1$ And we are done

Prove that for positive real numbers $a,b,c$ satisfying $abc=1$ the following inequality holds: $$ \frac{a^2}{b(1+c)} +\frac{b^2}{c(1+a)}+\frac{c^2}{a(1+b)} \ge \frac{3}{2} $$$$ \frac{a}{b(1+c^2)} +\frac{b}{c(1+a^2)}+\frac{c}{a(1+b^2)} \ge \frac{3}{2} $$

Titu's lemma super hawt

edit: exactly the same as mathluis :/ sorry, not intentional

Notice that if we rewire the inequality we obtain $\sum_{cyc}\frac{a}{b(1+c)}=\sum_{cyc}\frac{a^2}{ab(1+c)}=\sum_{cyc}\frac{a^2}{ab+1}\overset{\text{Engel C-S}}{\ge}\frac{\left(\sum_{cyc}a\right)^2}{\sum_{cyc}ab+3}$ Thus the inequality transforms into $2\left(\sum_{cyc}a\right)^2\ge3\sum_{cyc}ab+9\Longleftrightarrow 2\sum_{cyc}a^2+4\sum_{cyc}ab\ge3\sum_{cyc}ab+9\Longrightarrow 2\sum_{cyc}a^2+\sum_{cyc}ab\ge9$ Which is clearly true by $\text{AM-GM}$ since $\sum_{cyc}a^2\ge3\sqrt[3]{a^2b^2c^2}=3\text{ and }\sum_{cyc}ab\ge3\sqrt[3]{a^2b^2c^2}=3$ $\blacksquare$.