<s>[INST] If $xyz=32$, find the minimal value of If $xyz=32;x,y,z>0$, find the minimal value of $f(x,y,z)=x^2+4xy+4y^2+2z^2$
I tried to do by $A.M.\geq M.G.$:
$\frac{x^2+4y^2+2z^2}{2}\geq\sqrt{8x^2y^2z^2}\to x^2+4y^2+2z^2\geq32$
But how can I maximaze 4xy?
 [/INST] The  Alexey Burdin's hint gives the following substitution.
Let $x=4a$, $y=2b$ and $z=4c$.
Thus, $abc=1$ and by AM-GM we obtain:
$$x^2+4xy+4y^2+2z^2=16a^2+32ab+16b^2+32c^2=$$
$$=16(a^2+2ab+b^2+2c^2)\geq16\cdot6\sqrt[6]{a^4b^4c^4}=96.$$
The equality occurs for $a=b=c=1,$ which says that we got a minimal value.
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