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: Apply $AM \ge GM$ not to $8a^2 + b^2 + c^2$, but to $$ (2a)^2 + (2a)^2 + b^2 + c^2 $$ and $HM \le GM$ not to $a^{-1}+b^{-1}+c^{-1}$, but to $$ \frac{1}{2a} + \frac{1}{2a} + \frac{1}{b} + \frac{1}{c} $$
The “partitions” are chosen in such a way that equality can hold simultaneously in both estimates, in this case when $2a=b=c$.
One can also use the relationships between generalized means, here “harmonic mean $\le$ quadratic mean”: $$ \frac{4}{\frac{1}{2a} + \frac{1}{2a} + \frac{1}{b} + \frac{1}{c}} \le \sqrt{\frac{(2a)^2 + (2a)^2 + b^2 + c^2}{4}} $$
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.
