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 
 A: by the Cauchy-Schwarz inequality,we have
$$(a^2+64)(1+4)\ge(a+16)^2$$
$$(b^2+1)(4+1)\ge (2b+1)^2$$
then
$$\sqrt{a^2+64}+\sqrt{b^2+1}\ge\dfrac{1}{\sqrt{5}}(a+2b+17)$$
and By AM-GM,we have
$$a+2b\ge 2\sqrt{2ab}=8$$
so
$$\sqrt{a^2+64}+\sqrt{b^2+1}\ge 5\sqrt{5}$$
A: 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
$$
(1+\frac{8}{b})^2(1+b^2)-125=\frac{(b-2)^2\bigg(b^2+20b+16\bigg)}{b^2}
$$
A: 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.
