minimum value of $(8a^2+b^2+c^2)\cdot (a^{-1}+b^{-1}+c^{-1})^2$ 
If $a,b,c>0.$ Then minimum value of
$(8a^2+b^2+c^2)\cdot (a^{-1}+b^{-1}+c^{-1})^2$

Try: Arithmetic geometric inequality
$8a^2+b^2+c^2\geq 3\cdot 2\sqrt{2}(abc)^{1/3}$
and $(a^{-1}+b^{-1}+c^{-1})\geq 3(abc)^{-1/3}$
so $(8a^2+b^2+c^2)\cdot (a^{-1}+b^{-1}+c^{-1})^2\geq 18\sqrt{2}(abc)^{-1/3}$
could some help me to solve it.  answer is $64$

Hint: Another way is to consider Hölder's Inequality
$$(8a^2+b^2+c^2)(a^{-1}+b^{-1}+c^{-1})^2 \geqslant (2+1+1)^3$$
Equality is possible when $8a^3=b^3=c^3=1$ (why?), so this is the minimum.
