Find minimum $|z|$ satisfying $|z + 1/z |= 2$. When I tried this using normal complex inequalities like $|z_{1} - z_{2}| \ge ||z_{1}| - |z_{2}||$. $\sqrt 2 - 1$ came up but the real answer seems to be $(3 - 2\sqrt 2)^{1/2}$. Some online answers on other sites support my answer as well, but I am confused which ones correct. If the latter is correct please explain how.

$|z^2+1| = 2|z|$. Let $z = a+bi \implies |z^2+1| = |(a+bi)^2+1|=|a^2+2abi-b^2+1|=\sqrt{(a^2-b^2+1)^2+4a^2b^2}=2\sqrt{a^2+b^2}\implies (a^2-b^2+1)^2+4a^2b^2=4a^2+4b^2\implies a^4+b^4+1-2a^2b^2+2a^2-2b^2+4a^2b^2=4a^2+4b^2\implies a^4+b^4+1+2a^2b^2-2a^2-2b^2=4b^2\implies (a^2+b^2-1)^2=4b^2\le 4(a^2+b^2)$. Let $c = a^2+b^2=|z|^2\implies (c-1)^2 \le 4c \implies c^2-2c+1 - 4c \le 0\implies c^2-6c+1\le 0\implies (c-3)^2\le 8\implies |c-3| \le 2\sqrt{2}\implies -2\sqrt{2} \le c - 3\le 2\sqrt{2}\implies c \ge 3 - 2\sqrt{2}$. Thus $|z|_{\text{min}}=\sqrt{c_{\text{min}}}=\sqrt{3-2\sqrt{2}}$. This is achieved when $a = 0, b^2-1 = \pm 2b\implies a = 0, (b\pm1)^2=2\implies a = 0, |b\pm 1| = \sqrt{2}\implies a = 0, b\pm 1=\pm \sqrt{2}\implies a = 0, b = \mp1 \pm \sqrt{2}$
