By the AM-GM inequality for eleven numbers (or weighted AM-GM) we get $a + 10b \ge 11\sqrt[11]{ab^{10}}$. Similarly, we get
$$b + 22c \ge 23\sqrt[23]{bc^{22}} \quad \text{and} \quad c + 7a \ge 8\sqrt[8]{ca^7}.$$
Multiplying these three inequalities gives
$$(a + 10b)(b + 22c)(c + 7a) \ge 2024 a^{\frac{1}{11} + \frac{7}{8}} b^{\frac{10}{11} + \frac{1}{23}} c^{\frac{22}{23} + \frac{1}{8}}.$$
We will be finished if we can prove that
$$a^{\frac{1}{11} + \frac{7}{8}} b^{\frac{10}{11} + \frac{1}{23}} c^{\frac{22}{23} + \frac{1}{8}} \ge abc.$$
Since $\frac{1}{11} + \frac{7}{8} = \frac{1955}{2024}$, $\frac{10}{11} + \frac{1}{23} = \frac{1928}{2024}$ and $\frac{22}{23} + \frac{1}{8} = \frac{2189}{2024}$, this can be rewritten as
$$a^{1955} b^{1928} c^{2189} \ge a^{2024} b^{2024} c^{2024}, \text{ which simplifies to } c^{165} = c^{69} c^{96} \ge a^{69} b^{96}.$$
Since $c \ge a$ and $c \ge b$, we have $c^{69} \ge a^{69}$ and $c^{96} \ge b^{96}$, hence $c^{69}c^{96} \ge a^{69}b^{96}$, which concludes the proof.

Define $T = (a + 10b)(b + 22c)(c + 7a) - 2024abc$ and expand
$$T = 220c^2b + 22ac^2 + 10cb^2 - 483abc + 154a^2c + 70ab^2 + 7a^2b.$$
Define $x = c/a$ and $y = c/b$ and write $T/c^3$ in terms of $x$ and $y$ to get $T = \frac{c^3}{x^2y^2}S$, where $S = 220x^2y + 22xy^2 + 10x^2 - 483xy + 154y^2 + 70x + 7y$. Since $x \ge 1$ and $y \ge 1$, we can write $x = u + 1$ and $y = v + 1$ where $u, v \ge 0$ and obtain
$$S = 220u^2v + 22uv^2 + 230u^2 + uv + 176v^2 + 69u + 96v.$$
Clearly $S \ge 0$ and hence $T \ge 0$ as required.

After multiplying the terms on the left hand side and simplifying, the original inequality becomes
$$220c^2b + 22ac^2 + 10cb^2 + 154a^2c + 70ab^2 + 7a^2b \geq 483abc.$$
The sum of the coefficients on the left hand side is 483, hence we can use weighted AM-GM to obtain
$$\begin{aligned} & 220c^2b + 22ac^2 + 10cb^2 + 154a^2c + 70ab^2 + 7a^2b \\ & \geq 483(c^2b)^{\frac{220}{483}} (ac^2)^{\frac{22}{483}} (cb^2)^{\frac{10}{483}} (a^2c)^{\frac{154}{483}} (ab^2)^{\frac{70}{483}} (a^2b)^{\frac{7}{483}} \\ & = 483a^{\frac{414}{483}} b^{\frac{387}{483}} c^{\frac{648}{483}} \geq 483abc. \end{aligned}$$
The last inequality was obtained from
$$c^{\frac{69}{483}} \geq a^{\frac{69}{483}} \quad c^{\frac{96}{483}} \geq b^{\frac{96}{483}},$$
using $648 - 69 - 96 = 483$, $414 + 69 = 483$ and $387 + 96 = 483$.