Pinko wrote: Prove that for $\forall$ $z\in \mathbb{C}$ the following inequality is true: $|z|^2+2|z-1|\geq 1+2|z-1|\geq 1$, where $"="$ is reached when $z=1$. Wrong. Choose as counter-example $z=0$ and first inequality is wrong.

@pco, I fixed it, thanks.

$\left|z\right|^{2}+2\left|z-1\right|\ge\left|z\right|^{2}+2-2\left|z\right|\ge\left(2\left|z\right|-1\right)+2-2\left|z\right|=1$

If $z = a+bi$, we want $a^2 + b^2 + 2\sqrt{(a-1)^2 + b^2} \geq 1$. By crudely using $b\geq 0$, the left-hand side is at least $a^2 + 2|a-1| \geq 1$ and in the latter $a\geq 1$ gives the equivalent $(a-1)(a+3) \geq 0$ (with equality iff $a=1$, $b=0$, i.e. $z=1$) while $a<1$ gives $(a-1)^2 \geq 0$ and equality does not hold.