Inequality in complex numbers

Question

Prove that for all $z\in \mathbb{C}$ $$\frac{\Vert z+i\Vert z\Vert \Vert}{\Vert z+1\Vert}\leq \frac{2\Vert z \Vert}{\Vert z \Vert +1}$$

Answer 1

Let $z = r ( \cos t + i \sin t)$. Then the inequality says that

$$r \frac{\sqrt{ 2 + 2 \cos t}}{\sqrt{r^2 + 2r \cos t + 1}} \leq \frac{2r}{r+1}\ .$$

When this is squared and multiplied out, there is a factor of 1 - cos t which cancels, leaving the condition

$$r^2 - 2 r + 1 > 0\ .$$

Answer 2

The inequality is wrong

if $z=i$, then

$$\frac{\Vert z+i\Vert z\Vert \Vert}{\Vert z+1\Vert}= \frac{\Vert i+i\vert\vert}{\Vert i+1\Vert}=\sqrt2$$

but

$$\frac{2\Vert z \Vert}{\Vert z \Vert +1}= \frac{2}{\Vert i\Vert +1}=1$$