# Inequlity

• Jan 24th 2010, 12:58 PM
ns1954
Inequlity
Let $\displaystyle a>0$ and $\displaystyle k>1$. Prove that for comlex $\displaystyle z$

$\displaystyle |arg\ (z)|\leq 2\,\arccos\frac 1k \Rightarrow |a+z|\geq \frac {a+|z|}k$

Hint: $\displaystyle \cos x$ is decreasind in $\displaystyle [o,\pi]$. Not so difficult, I think?
• Aug 20th 2010, 02:17 AM
halbard
Inequality

Let $\displaystyle \theta=\arg z$. Since $\displaystyle |\arg z|\leq 2\arccos\dfrac1k=2\alpha$ and $\displaystyle \cos x$ is decreasing on $\displaystyle [0,\pi]$ we have

$\displaystyle \cos\theta\geq\cos 2\alpha=2\cos^2\alpha-1=\dfrac2{k^2}-1$.

Thus

$\displaystyle \mathop{\textrm{Re}} z=|z|\cos\theta\geq\left(\dfrac2{k^2}-1\right)|z|$.

Now
$\displaystyle |a+z|^2=a^2+2a\mathop{\textrm{Re}}z+|z|^2\geq a^2+2a|z|\left(\dfrac2{k^2}-1\right)+|z|^2$

Hence
$\displaystyle |a+z|^2-\dfrac{(a+|z|)^2}{k^2}\geq\left(1-\dfrac1{k^2}\right)(a^2-2a|z|+|z|^2)$

Since $\displaystyle a^2-2a|z|+|z|^2=(a-|z|)^2\geq 0$ and $\displaystyle k>1$ the result follows.