From the given condition, we have $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ By weighted AM-GM, $$\frac{1}{a}\cdot\frac{bc}{a^a}+\frac{1}{b}\cdot\frac{ca}{b^b}+\frac{1}{c}\cdot\frac{ab}{c^c}\geq\prod_{cyc}\frac{(bc)^{\frac{1}{a}}}{a}=\prod_{cyc}\frac{1}{a^{\frac{1}{a}}}$$But, again by weighted AM-GM, we have $$\frac{1}{a}\cdot a+\frac{1}{b}\cdot b+\frac{1}{c}\cdot c\geq\prod_{cyc}a^{\frac{1}{a}}$$So, $$\frac{1}{\prod_{cyc}a^{\frac{1}{a}}}\geq\frac{1}{3}$$, and hence, we are done $\blacksquare$

parmenides51 wrote: Let $a, b, c$ be positive real numbers with $ab + bc + ac = abc$. Prove that $$\frac{bc}{a^{a+1}} +\frac{ac}{b^{b+1 }}+\frac{ab}{c^{c+1}} \ge \frac13$$ https://cms.math.ca/Competitions/REP/2020/2020CMOQR_exam_en.pdf

Substituting $a,b,c$ with $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$ we have $a+b+c=1$ and $$\sum_{cyc}a^{\frac{1}{a}+2}\geq \dfrac{abc}{3}$$Given $f(x)=x^{\frac{1}{x}+2}$ $$f''(x)=x^{\frac{1}{x}+2}(\ln^2(x)-2(x+1)\ln(x)+2x^2+x+1)\geq x^{\frac{1}{x}+2}(\ln^2(x)-2(x+1)(x-1)+2x^2+x+1)=x^{\frac{1}{x}+2}(\ln^2(x)+x+3)\geq 0$$So by Jensen + $AM-GM$ we get $$\sum_{cyc}a^{\frac{1}{a}+2}\geq \dfrac{1}{3^4}\geq \dfrac{abc}{3}$$