By the Cauchy-Schwarz inequality, we have
$$3\sqrt{1 + 5a^2} = \sqrt{2^2 + (\sqrt{5})^2} \cdot \sqrt{1 + 5a^2} \geq 2 + 5a$$
and
$$3\sqrt{2 + b^2} = \sqrt{(2\sqrt{2})^2 + 1} \cdot \sqrt{2 + b^2} \geq 4 + b.$$
Hence
$$\sqrt{1 + 5a^2} + 5\sqrt{2 + b^2} \geq \frac{2 + 5a}{3} + \frac{5(4 + b)}{3} = 9.$$
Equality holds for $a = b = 1/2$.