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

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  • $\begingroup$ Hint: $|\Im(z)|\leq||z||$ and $f(x) = \frac{2x}{x+1}$ is an increasing function since $f'(x) = \frac{2}{(x+1)^2} > 0$. $\endgroup$ – Chris K Jun 15 '14 at 19:33
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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

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

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  • $\begingroup$ Please: on the left in the numerator cos t OR sin t $\endgroup$ – mohd Sep 25 '14 at 16:32
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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

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

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