2. Given positive numbers $a, b, c, d$. Find the minimum value of the expression

$$
A=\left(a+\frac{1}{b}\right)^{3}+\left(b+\frac{1}{c}\right)^{3}+\left(c+\frac{1}{d}\right)^{3}+\left(d+\frac{1}{a}\right)^{3}
$$
Answer: 32.

Solution 1. We will use the Cauchy inequality for means first in each parenthesis, and then for the entire sum. We get

$$
A \geqslant\left(2 \sqrt{\frac{a}{b}}\right)^{3}+\left(2 \sqrt{\frac{b}{c}}\right)^{3}+\left(2 \sqrt{\frac{c}{d}}\right)^{3}+\left(2 \sqrt{\frac{d}{a}}\right)^{3} \geqslant 32\left(\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{d} \cdot \frac{d}{a}\right)^{3 / 8}=32
$$

Equality is achieved when $a=b=c=d=1$.

Solution 2. We will use the inequality $x^{3}+y^{3} \geqslant \frac{1}{4}(x+y)^{3}$, which is true for $x, y>0$. Applying it three times, we get

$$
x^{3}+y^{3}+z^{3}+w^{3} \geqslant \frac{1}{4}\left((x+y)^{3}+(z+w)^{3}\right) \geqslant \frac{1}{16}(x+y+z+w)^{3} \quad \text { for } \quad x, y, z, w>0
$$

Then, by the Cauchy inequality for means,

$$
A \geqslant \frac{1}{16}\left(a+\frac{1}{b}+b+\frac{1}{c}+c+\frac{1}{d}+d+\frac{1}{a}\right)^{3} \geqslant \frac{1}{16}\left(8 \sqrt[8]{a \cdot \frac{1}{b} \cdot b \cdot \frac{1}{c} \cdot c \cdot \frac{1}{d} \cdot d \cdot \frac{1}{a}}\right)^{3}=\frac{8^{3}}{16}=32
$$

Equality is achieved when $a=b=c=d=1$.