I solved the following question using AM-GM inequality while practicing in an app
"If $x,y,z$ are positive real numbers satisfying $xyz=32$, find the minimum value of:
$$x^2+4xy+4y^2+2z^2$$
In the question discussion, many people commented that we must show equality exists for the maxima/minima.
Why do we need to?
Sorry for my poor english. Thank you.
By AM-GM twice we obtain: $$x^2+4xy+4y^2+2z^2=(x+2y)^2+2z^2\geq(2\sqrt{2xy})^2+2z^2=8xy+2z^2=$$ $$=4xy+4xy+2z^2\geq3\sqrt{(4xy)^2\cdot2z^2}=96.$$ The equality occurs for $x=2y$, $2xy=z^2$ and $xyz=32$ or $(x,y,z)=(4,2,4),$ which says that we got a minimal value.
Indeed, $$4^2+4\cdot4\cdot2+4\cdot2^2+2\cdot4^2=96,$$ which is a minimal mvalue.