66th Belarusian Mathematical Olympiad

(Solution by S. Shaban, A. Vasileuski) By the Cauchy-Bunyakovsky-Schwarz inequality, $$\sqrt{a^2+\frac{c^2}{b^2}}\sqrt{b^2+\frac{a^2}{c^2}}\ge \left(ab+\frac{a}{b}\right)=a\left(b+\frac{1}{b}\right),$$ $$\sqrt{b^2+\frac{a^2}{c^2}}\sqrt{c^2+\frac{b^2}{a^2}}\ge \left(bc+\frac{b}{c}\right)=b\left(c+\frac{1}{c}\right),$$ $$\sqrt{c^2+\frac{b^2}{a^2}}\sqrt{a^2+\frac{c^2}{b^2}}\ge \left(ca+\frac{c}{a}\right)=c\left(a+\frac{1}{a}\right).$$ Multiplying these inequalities, we obtain the required inequality.