The techniques used here solves this problem too. Given inequality is equivalent to $$\sum_{cyc} \dfrac{a^2 c}{a+b} \geq \dfrac{3}{2}$$ Now by taking $\alpha_i=\dfrac{a}{\sqrt{a+b}}$ and $\beta_i=\sqrt{a+b}$ ; other terms similarly and applying Cauchy Schwarz, we get our result. I hope this works

Given inequality is equivalent to $$\sum_{cyc} \dfrac{a^2 c}{a+b} \geq \dfrac{3}{2}$$. $$\sum_{cyc} \dfrac{a^2 c}{a+b}\sum_{cyc}c(a+b)\ge (\sum_{}^{}ac)^2 \rightarrow \sum_{cyc} \dfrac{a^2 c}{a+b} \geq \frac{\sum_{}^{}ac}{2}\geq \frac{3}{2}$$

We get $$\sum_{cyc} \dfrac{a^2 c}{a+b} \geq \dfrac{3}{2}$$$$\sum_{cyc} \dfrac{a^2 c^2}{c(a+b)} \geq \dfrac{3}{2}$$By Titu's lema $$\sum_{cyc} \dfrac{a^2 c^2}{c(a+b)} \geq \frac{(ab+bc+ac)^2}{2(ab+bc+ac)}$$$\frac{(ab+bc+ac)^2}{2(ab+bc+ac)} = \frac{ab+bc+ac}{2}$ So we want to prove that $ab+bc+ac \geq 3$ which is obvious by AM-GM

parmenides51 wrote: Prove that for positive real numbers $a, b, c$ such that $abc=1$, the following inequality holds: $$\frac{a}{b(a+b)}+\frac{b}{c(b+c)}+\frac{c}{a(c+a)} \ge \frac32$$ (I. Voronovich) Prove that for positive real numbers $a, b, c$ , the following inequality holds: $$\frac{a}{b(a+b)}+\frac{b}{c(b+c)}+\frac{c}{a(c+a)} \ge \frac{3}{2\sqrt[3]{abc}}$$

parmenides51 wrote: Prove that for positive real numbers $a, b, c$ such that $abc=1$, the following inequality holds: $$\frac{a}{b(a+b)}+\frac{b}{c(b+c)}+\frac{c}{a(c+a)} \ge \frac32$$ (I. Voronovich) See also here https://artofproblemsolving.com/community/c6h1219608p13447063

$$\sum_{cyc}\frac{a}{b(a+b)}=\sum_{cyc}\frac{ca^2}{a+b}=\sum_{cyc}\frac{c^2a^2}{c(a+b)}\overset{\text{Engel C-S}}{\ge}\frac{\left(\sum_{cyc}ab\right)^2}{2\sum_{cyc}ab}=\frac{\sum_{cyc}ab}{2}\ge\frac{3}{2}$$$$\blacksquare.$$