<s>[INST] Minimum value of magnitude of complex expression 
If $|2z_1 + \bar z_2| = 2\sqrt2$ and   $|1 + 2z_1z_2 | = 3 $ then minimum value of $(|z_1|^2 + 4|z_2|^2)$, is:-

Now
$|2z_1 + \bar z_2|^2 = 8$
$(2 z_1 + \bar z_2)(2\bar z_1 + z_2) = 8$
$4z_1\bar z_1 + 2 z_1z_2 + 2\bar z_1\bar z_2 + z_2\bar z_2 = 8$

$4|z_1|^2 + |z_2|^2 + 2 z_1z_2 + 2\bar z_1\bar z_2 = 8.......(1)$

Now
$|1 + 2z_1z_2 |^2 = 9 $
$(1 + 2z_1z_2)(1 + 2\bar z_1\bar z_2) = 9 $

$1 + 4|z_1|^2|z_2|^2 + 2z_1z_2 + 2\bar z_1\bar z_2= 9........(2)$

$(1) - (2)$
$4|z_1|^2 + |z_2|^2 = 4|z_1|^2|z_2|^2$
I am not able to make any progress after this step.
 [/INST] Let $|z_1|^2 = x$, $|z_2|^2 = y$. Then we have to find the minimum value of $x+4y$. Continuing from where you left off:
$$4x + y = 4xy \Rightarrow 4xy-y = 4x \Rightarrow y = \frac{4x}{4x-1}$$
and so:
$$x+4y=x + \frac{16x}{4x-1} = x + \frac{16x-4}{4x-1} + \frac{4}{4x-1} = x+4+\frac{4}{4x-1}$$
By AM-GM, $(x-\frac{1}{4})+\frac{4}{4x-1} ≥ 2 \sqrt{(x-\frac{1}{4}) \cdot \frac{4}{4x-1}} = 2$ since $x,y ≥ 0$.
Hence, the minimum value of $x + 4y = x - \frac{1}{4} + \frac{4}{4x-1} + \frac{17}{4}$ is $2 + \frac{17}{4} = \boxed{\frac{25}{4}}$.
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