Let $r\in[0,1]$ and $z\in\mathbb C,\mathrm{Im}z> 0.$ I am struggling to prove the following inequality: $$\frac{2(1+|z|)r}{\sqrt{|z^2-4|}}\wedge\sqrt{(1+|z|)r}\leq3\left(\left[\left(1+\frac{1}{\sqrt{|z^2-4|}}\right)r\right]\wedge\sqrt{r}\right),$$ where $x\wedge y=\min(x,y).$ This is trivial for $r=0.$ First I supposed that $r>0$ and $$\frac{2(1+|z|)r}{\sqrt{|z^2-4|}}\leq\sqrt{(1+|z|)r},$$ which implies that $$\sqrt{r}\leq\frac{\sqrt{|z^2-4|}}{2\sqrt{(1+|z|)}},$$ but I was unable to get something non-trivial out of this.

  • $\begingroup$ What happens when $z=2$? $\endgroup$ – Sagi Shadur Jul 12 '16 at 18:56
  • $\begingroup$ It holds since $\sqrt{3r}\leq 3 sqrt{r}$ $\endgroup$ – themightymoose Jul 12 '16 at 18:59
  • $\begingroup$ so you consider the devision by zero as infinity? $\endgroup$ – Sagi Shadur Jul 12 '16 at 19:01
  • $\begingroup$ The limit at 2 is +infinity and there is a minimum anyway, so both sides are well defined at 2 $\endgroup$ – themightymoose Jul 12 '16 at 19:04
  • $\begingroup$ First of all, You can simplify the inequality. pay attension that if $a,b,c\ge0$ then $a \min{(b,c)}=\min{(ab,ac)}$. Now, since $\sqrt{|z^2-4|}\ge0$ we can multiply/divide the inequality with this term and the result won't change. therefore the inequality is the same as: $2(1+|z|)r \wedge \sqrt{(1+|z|)r}\cdot\sqrt{|z^2-4|} \leq 3\left(\left(\sqrt{|z^2-4|}+1\right)r \wedge \sqrt{|z^2-4|}\cdot\sqrt{r}\right)$ $2r(1+|z|) \wedge \sqrt{(1+|z|)}\cdot\sqrt{r|z^2-4|} \leq 3r\left(\sqrt{|z^2-4|}+1\right) \wedge 3\sqrt{r|z^2-4|}$ $\endgroup$ – Sagi Shadur Jul 12 '16 at 19:21

It is not true.

For $r=0.01,z=2+0.1i$, we have $$|z^2-4|=|-0.01+0.4i|=\sqrt{0.1601},\quad |z|=\sqrt{4.01}$$ $$\frac{2(1+|z|)r}{\sqrt{|z^2-4|}}=\frac{2(1+\sqrt{4.01})0.01}{\sqrt{\sqrt{0.1601}}}\approx 0.095$$

$$\sqrt{(1+|z|)r}=\sqrt{(1+\sqrt{4.01})0.01}\approx 0.173$$

$$3\left[\left(1+\frac{1}{\sqrt{|z^2-4|}}\right)r\right]=3\left[\left(1+\frac{1}{\sqrt{\sqrt{0.1601}}}\right)0.01\right]\approx 0.077$$


Therefore, the following inequality does not hold : $$\frac{2(1+|z|)r}{\sqrt{|z^2-4|}}\wedge\sqrt{(1+|z|)r}\leq 3\left[\left(1+\frac{1}{\sqrt{|z^2-4|}}\right)r\right]\wedge 3\sqrt{r}$$


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