By rearranging the left hand side of the inequality we get
$$\begin{aligned} \left(1 + \frac{c}{a}\right) \left(1 + \frac{c}{b}\right) &= \frac{(a+c)(b+c)}{ab} = \frac{c^2 + ab + c(a+b)}{ab} \\ &= \frac{c^2}{ab} + 1 + \frac{c(a+b)}{ab} \end{aligned}$$
Since the triangle is right it follows that $a^2 + b^2 = c^2$
$$= \frac{a^2 + b^2}{ab} + 1 + \frac{\sqrt{a^2 + b^2}(a+b)}{ab}$$

Now we use $a^2 + b^2 \ge 2ab$ and $a + b \ge 2\sqrt{ab}$ to get
$$\begin{aligned} \ge & \frac{2ab}{ab} + 1 + \frac{\sqrt{2ab} \cdot 2\sqrt{ab}}{ab} \\ = & 2 + 1 + 2\sqrt{2} = 3 + 2\sqrt{2}. \end{aligned}$$