1. **Problem statement:** We want to find the minimum value of $$S = \frac{12 x^3}{y^2} + \frac{y}{x^2} + 3y$$ for positive variables $x, y > 0$, using the AM-GM inequality. 2. **Recall the AM-GM inequality:** For positive numbers $a_1, a_2, \dots, a_n$, $$\frac{a_1 + a_2 + \cdots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \cdots a_n}$$ with equality if and only if $a_1 = a_2 = \cdots = a_n$. 3. **Apply AM-GM to the three terms:** Let $$a = \frac{12 x^3}{y^2}, \quad b = \frac{y}{x^2}, \quad c = 3y.$$ By AM-GM, $$\frac{a + b + c}{3} \geq \sqrt[3]{abc}.$$ So, $$S = a + b + c \geq 3 \sqrt[3]{abc}.$$ 4. **Calculate the product $abc$:** $$abc = \frac{12 x^3}{y^2} \times \frac{y}{x^2} \times 3y = 12 \times 3 \times \frac{x^3}{y^2} \times \frac{y}{x^2} \times y = 36 \times \frac{x^{3-2}}{y^{2-1-1}} = 36 x^{1} y^{0} = 36 x.$$ 5. **So the inequality becomes:** $$S \geq 3 \sqrt[3]{36 x} = 3 \times 36^{1/3} x^{1/3}.$$ 6. **We want to minimize $S$ over $x,y>0$.** From step 5, $S$ depends on $x$ as $$S \geq 3 \times 36^{1/3} x^{1/3}.$$ To minimize $S$, minimize $x^{1/3}$, but $x > 0$ and also recall from step 3 that equality holds when $$a = b = c,$$ which means $$\frac{12 x^3}{y^2} = \frac{y}{x^2} = 3y.$$ 7. **Set $\frac{y}{x^2} = 3y$:** Since $y > 0$, divide both sides by $y$: $$\frac{1}{x^2} = 3 \implies x^2 = \frac{1}{3} \implies x = \frac{1}{\sqrt{3}}.$$ 8. **Set $\frac{12 x^3}{y^2} = 3y$:** Substitute $x = \frac{1}{\sqrt{3}}$: $$\frac{12 \left(\frac{1}{\sqrt{3}}\right)^3}{y^2} = 3y.$$ Calculate numerator: $$\left(\frac{1}{\sqrt{3}}\right)^3 = \frac{1}{3 \sqrt{3}}.$$ So, $$\frac{12}{3 \sqrt{3} y^2} = 3y \implies \frac{12}{3 \sqrt{3} y^2} = 3y.$$ Multiply both sides by $y^2$: $$\frac{12}{3 \sqrt{3}} = 3 y^3.$$ Simplify left side: $$\frac{12}{3 \sqrt{3}} = \frac{4}{\sqrt{3}}.$$ So, $$3 y^3 = \frac{4}{\sqrt{3}} \implies y^3 = \frac{4}{3 \sqrt{3}}.$$ 9. **Solve for $y$:** $$y = \sqrt[3]{\frac{4}{3 \sqrt{3}}}.$$ 10. **Calculate minimum $S$:** Recall from step 3 that at equality, $$S = 3a = 3b = 3c.$$ Use $c = 3y$: $$S_{min} = 3 \times 3y = 9y = 9 \times \sqrt[3]{\frac{4}{3 \sqrt{3}}}.$$ **Final answers:** $$x = \frac{1}{\sqrt{3}}, \quad y = \sqrt[3]{\frac{4}{3 \sqrt{3}}}, \quad S_{min} = 9 \sqrt[3]{\frac{4}{3 \sqrt{3}}}.$$