<s>[INST] 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 
 [/INST] 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
$$\sqrt{a^2+64}+\sqrt{b^2+1}\ge 5\sqrt{5}$$
</s>