First solution. The inequality can be written
$$\frac{1}{a} + \frac{3}{b} + \frac{5}{c} + b^2 + 2c^2 \geq 4(a^2 + b^2 + c^2)$$
or
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{2}{b} + b^2 + \frac{4}{c} + 2c^2 \geq 12.$$
This follows from the following inequalities by using the AM-GM inequality:
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq \frac{3}{\sqrt[3]{abc}} \geq 3$$
because
$$3 = a^2 + b^2 + c^2 \geq 3\sqrt[3]{a^2b^2c^2} \implies abc \leq 1.$$
On the other hand,
$$\frac{2}{b} + b^2 = \frac{1}{b} + \frac{1}{b} + b^2 \geq 3\sqrt[3]{\frac{1}{b} \cdot \frac{1}{b} \cdot b^2} = 3$$
$$\frac{4}{c} + 2c^2 = \frac{2}{c} + \frac{2}{c} + 2c^2 \geq 3\sqrt[3]{\frac{2}{c} \cdot \frac{2}{c} \cdot 2b^2} = 6.$$
The equality holds if $a = b = c = 1$.

Second solution. As $x^3 - 3x + 2 \geq 0, \forall x \geq 0$ (equivalent to $(x - 1)^2(x + 2) \geq 0$; alternatively, the previous inequality follows from AM-GM: $x^3 + 1 + 1 \geq 3\sqrt[3]{x^3 \cdot 1 \cdot 1} = 3x$; the equality holds whence $x = 1$). We deduce that $\frac{1}{x} \geq \frac{3}{2} - \frac{x^2}{2}$. Writing this for $a$, $b$, $c$ and multiplying these inequalities by 1, 3, and 5, respectively, we obtain by summing $\frac{1}{a} + \frac{3}{b} + \frac{5}{c} \geq \frac{27}{2} - \frac{a^2 + 3b^3 + 5c^2}{2}$. Using $27 = 9(a^2+b^2+c^2)$ one obtains the conclusion.
The equality holds if and only if $a = b = c = 1$.