Using the AM-GM inequality for three terms twice, one gets
$$\begin{aligned} \frac{2c}{a} + \frac{c}{b} &= \frac{(2b+a)c}{ab} = \frac{(2b+a)\sqrt{2a^2+b^2}}{3ab} = \frac{(b+b+a)\sqrt{a^2+a^2+b^2}}{3ab} \\ &\ge \frac{3\sqrt[3]{b^2a}\sqrt{3\sqrt[3]{a^4b^2}}}{3ab} = \frac{3\sqrt{3}\sqrt[3]{b^2a \cdot a^2b}}{3ab} = \frac{3\sqrt{3}ab}{3ab} = \sqrt{3}. \end{aligned}$$

Using HM-QM inequality for $a, a, b$ gives
$$\frac{3}{\frac{2}{a} + \frac{1}{b}} = \frac{3}{\frac{1}{a} + \frac{1}{a} + \frac{1}{b}} \le \sqrt{\frac{a^2 + a^2 + b^2}{3}} = \sqrt{\frac{2a^2 + b^2}{3}} = \sqrt{\frac{9c^2}{3}} = \sqrt{3}c.$$
Thus
$$\left(\frac{2}{a} + \frac{1}{b}\right) \cdot c \ge \frac{3}{\sqrt{3}} = \sqrt{3},$$
which implies the desired inequality.

$$\begin{aligned} \left(\frac{2c}{a} + \frac{c}{b}\right)^2 &= \frac{(2b+a)^2 c^2}{a^2 b^2} = \frac{(2b+a)^2 (2a^2 + b^2)}{9a^2 b^2} = \\ &= \frac{((a^2 + b^2) + 4ab + 3b^2)(a^2 + (a^2 + b^2))}{9a^2 b^2} \ge \frac{(6ab + 3b^2)(a^2 + 2ab)}{9a^2 b^2} = \\ &= \frac{(2a+b)(a+2b)}{3ab} = \frac{2a^2 + 5ab + 2b^2}{3ab} \ge \frac{4ab + 5ab}{3ab} = 3. \end{aligned}$$

The square of the l.h.s. of the desired inequality is
$$\left(\frac{2c}{a} + \frac{c}{b}\right)^2 = \frac{(2b+a)^2(2a^2+b^2)}{9a^2b^2} = \frac{1}{9}\left(2+\frac{a}{b}\right)^2\left(2+\frac{b^2}{a^2}\right).$$
Denoting $\frac{a}{b} = x$, the desired inequality reduces to $\frac{1}{9}(2+x)^2(2+\frac{1}{x^2}) \ge 3$, which is equivalent to $(2+x)^2(2x^2+1) \ge 27x^2$. This in turn is equivalent to $2x^4+8x^3-18x^2+4x+4 \ge 0$, that is $2(x-1)^2(x^2+6x+2) \ge 0$ after factorization. This inequality holds, since on positive arguments the quadratic polynomial $x^2+6x+2$ is positive.