How find this inequality $\sqrt{a^2+64}+\sqrt{b^2+1}$ let $a,b$ are positive numbers,and such $ab=8$ find this minum

$$\sqrt{a^2+64}+\sqrt{b^2+1}$$

My try:

and I find when $a=4,b=2$,then
  $$\sqrt{a^2+64}+\sqrt{b^2+1}$$ is minum $5\sqrt{5}$

it maybe use Cauchy-Schwarz inequality 
Thank you 

Setting $x = a$ and $b = \frac{8}{x}$ we find that we want to minimise $$\sqrt{x^2 + 64} + \sqrt{\frac{64}{x^2} + 1} = \left (\frac{1}{x} + 1 \right ) \sqrt{x^2 + 64}$$
Differentiating, we obtain $$ \frac{x^3 - 64}{x^2 (x^2 + 64)} $$ Setting this equal to zero, we obtain a real extremum at $x=4$, corresponding to $a=4, b=2$ and a value of $5\sqrt{5}$. Analysis of the second derivative shows that this is a minimum.
