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 

As noted by Mher Safaryan, your inequality is equivalent to
$$
(1+a)\sqrt{b^2+1} \geq 5\sqrt{5} \tag{1}
$$
or in other words,
$$
(1+a)^2(1+b^2) \geq 125 \tag{2}
$$
or equivalently,
$$
(1+\frac{8}{b})^2(1+b^2) \geq 125 \tag{3}
$$
We are now done, because of the identity
$$
(1+\frac{8}{b})^2(1+b^2)-125=\frac{(b-2)^2\bigg(b^2+20b+16\bigg)}{b^2}
$$
