inequality proof

**lllll** · September 10th 2008, 11:06 PM

I'm a little stuck with the following inequality and was wondering if anyone could help me out.

$\frac{\mbox{Re}(z_1+z_2)}{|z_3+z_4|} \leq \frac{|z_1|+|z_2|}{||z_3|-|z_4||} \ \ \mbox{where} \ \ z_3 \neq z_4$

$\frac{x_1+x_2}{|z_3+z_4|} \leq \frac{|z_1|+|z_2|}{||z_3|-|z_4||}$

$\frac{x_1+x_2}{|z_3+z_4|} \leq \frac{\sqrt{x_1^2+y_1^2}+\sqrt{x_2^2+y_2^2}}{||z_3 |-|z_4||}$

at which point I get stuck.

I was also thinking of using the property that:

$|z_3+z_4| \geq ||z_3|-|z_4|| \therefore \frac{1}{|z_3+z_4|} \leq \frac{1}{||z_3|-|z_4||}$ but the numerators are throwing me off.

**Moo** · September 10th 2008, 11:49 PM

Hello,

I was also thinking of using the property that:

$|z_3+z_4| \geq ||z_3|-|z_4||$

What is this property ?

but the numerators are throwing me off.

By the numerators, were you thinking of $x_1+x_2 \le \sqrt{x_1^2+y_1^2}+\sqrt{x_2^2+y_2^2}$ ?
$\sqrt{x_1^2+y_1^2}+\sqrt{x_2^2+y_2^2} \ge \sqrt{x_1^2}+\sqrt{x_2^2}=|x_1|+|x_2| \ge x_1+x_2$

**Laurent** · September 11th 2008, 02:26 AM

For the numerator, use Moo's idea, and for the denominator, use indeed yours (the lower bound in the triangular inequality).

About this lower bound, it is a consequence of the usual triangle inequality: for any $x,y$ , $|x|\leq|x+y|+|-y|$ , hence $|x|-|y|\leq |x+y|$ and by symmetry the same holds swapping $x$ and $y$ , so that $||x|-|y||=\max(|x|-|y|,|y|-|x|)\leq |x+y|$ .

Laurent.

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