# Complex Inequality

• 10 Feb 2018, 12:32 AM
altSwift
Complex Inequality
Attachment 34449

Confused with iii). Any help would be appreciated!
• 10 Feb 2018, 12:43 AM
InteGrand
Re: Complex Inequality
Quote:

Originally Posted by altSwift
Attachment 34449

Confused with c). Any help would be appreciated!

$\noindent I assume you could do i) and ii). Here's some hints for iii). Note that if y=0 (note then x \neq 0 since z can't be 0), then z=x. So you just need to show that$

$\left|x + \frac{1}{x}\right| \geq 2$

$\noindent if x is real.$

$\noindent To show this, first show that x + \frac{1}{x} \geq 2 if x > 0. The result for negative x will follow immediately because \left|x + \frac{1}{x}\right| = |x| + \frac{1}{|x|} for \textbf{real} x(\neq 0).$

$\noindent So ultimately, it's enough to show that x + \frac{1}{x} \geq 2 if x > 0. Can you think of a way to show it? (Hint: rearrange and recall that perfect squares are non-negative!)$
• 10 Feb 2018, 1:09 AM
altSwift
Re: Complex Inequality
Quote:

Originally Posted by InteGrand
$\noindent I assume you could do i) and ii). Here's some hints for iii). Note that if y=0 (note then x \neq 0 since z can't be 0), then z=x. So you just need to show that$

$\left|x + \frac{1}{x}\right| \geq 2$

$\noindent if x is real.$

$\noindent To show this, first show that x + \frac{1}{x} \geq 2 if x > 0. The result for negative x will follow immediately because \left|x + \frac{1}{x}\right| = |x| + \frac{1}{|x|} for \textbf{real} x(\neq 0).$

$\noindent So ultimately, it's enough to show that x + \frac{1}{x} \geq 2 if x > 0. Can you think of a way to show it? (Hint: rearrange and recall that perfect squares are non-negative!)$

yeah sorry I meant confused with iii), will fix now and have another try
• 10 Feb 2018, 11:32 AM
altSwift
Re: Complex Inequality
Quote:

Originally Posted by InteGrand
$\noindent I assume you could do i) and ii). Here's some hints for iii). Note that if y=0 (note then x \neq 0 since z can't be 0), then z=x. So you just need to show that$

$\left|x + \frac{1}{x}\right| \geq 2$

$\noindent if x is real.$

$\noindent To show this, first show that x + \frac{1}{x} \geq 2 if x > 0. The result for negative x will follow immediately because \left|x + \frac{1}{x}\right| = |x| + \frac{1}{|x|} for \textbf{real} x(\neq 0).$

$\noindent So ultimately, it's enough to show that x + \frac{1}{x} \geq 2 if x > 0. Can you think of a way to show it? (Hint: rearrange and recall that perfect squares are non-negative!)$

$I mangaed to get this after lots of tries$

$\left|x + \frac{1}{x}\right| = \left|x + \frac{1}{x} -2 + 2\right|$

$= \left|\left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^{2} + 2\right|$

$since \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^{2} will always be \geq 0,$

$\left|x + \frac{1}{x}\right| \geq 2$

$therefore \left|z + \frac{1}{z}\right| \geq 2$

$\noindent It took me quite a while to come up with that, I don't think I would have the intuition to do that in a test without your help. Is there any alternative?$
• 10 Feb 2018, 12:22 PM
InteGrand
Re: Complex Inequality
Quote:

Originally Posted by altSwift
$I mangaed to get this after lots of tries$

$\left|x + \frac{1}{x}\right| = \left|x + \frac{1}{x} -2 + 2\right|$

$= \left|\left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^{2} + 2\right|$

$since \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^{2} will always be \geq 0,$

$\left|x + \frac{1}{x}\right| \geq 2$

$therefore \left|z + \frac{1}{z}\right| \geq 2$

$\noindent It took me quite a while to come up with that, I don't think I would have the intuition to do that in a test without your help. Is there any alternative?$

$\noindent Well if we want to show that$

$x + \frac{1}{x}\geq 2 \quad \text{if }x > 0,$

$\noindent then a perhaps more intuitive approach is to get everything on one side first. So it's equivalent to show that$

$x + \frac{1}{x} -2 \geq 0 \quad \text{if }x > 0.$

$\noindent Now you can either recognise the LHS as \left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2} straightaway, or if the \frac{1}{x} term is bothering you, you can multiply the inequality through by x to instead show the equivalent inequality$

$x^2 -2x + 1 \geq 0 \quad \text{if }x > 0.$

$\noindent Now you should be able to easily recognise the LHS as a perfect square. When writing the proof for a HSC exam though, you should probably write it up \emph{starting} from (x-1)^2 \geq 0 (x > 0) and then manipulating it into x +\frac{1}{x}\geq 2.$

$\noindent An alternative way to immediately see the result is via the \textbf{AM-GM inequality}, which is seen in the inequalities section in the Harder 3U topic.$
• 10 Feb 2018, 1:33 PM
altSwift
Re: Complex Inequality
Quote:

Originally Posted by InteGrand
$\noindent Well if we want to show that$

$x + \frac{1}{x}\geq 2 \quad \text{if }x > 0,$

$\noindent then a perhaps more intuitive approach is to get everything on one side first. So it's equivalent to show that$

$x + \frac{1}{x} -2 \geq 0 \quad \text{if }x > 0.$

$\noindent Now you can either recognise the LHS as \left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2} straightaway, or if the \frac{1}{x} term is bothering you, you can multiply the inequality through by x to instead show the equivalent inequality$

$x^2 -2x + 1 \geq 0 \quad \text{if }x > 0.$

$\noindent Now you should be able to easily recognise the LHS as a perfect square. When writing the proof for a HSC exam though, you should probably write it up \emph{starting} from (x-1)^2 \geq 0 (x > 0) and then manipulating it into x +\frac{1}{x}\geq 2.$

$\noindent An alternative way to immediately see the result is via the \textbf{AM-GM inequality}, which is seen in the inequalities section in the Harder 3U topic.$

$\noindent Ah I see now, starting from (x-1)^2 \geq 0 (x > 0) makes more sense because I always thought messing with inequalities instead of equating LHS and RHS wasn't allowed or isn't a real proof. Thank you.$

$\noindent I'll also look into the AM-GM inequality, although terry lee doesn't explicitly have it as a theorem. We didn't do GP and AP means in 3U, although there are exercises in cambridge on it. Is it required for 3U/4U?$
• 10 Feb 2018, 1:53 PM
InteGrand
Re: Complex Inequality
Quote:

Originally Posted by altSwift
$\noindent Ah I see now, starting from (x-1)^2 \geq 0 (x > 0) makes more sense because I always thought messing with inequalities instead of equating LHS and RHS wasn't allowed or isn't a real proof. Thank you.$

$\noindent I'll also look into the AM-GM inequality, although terry lee doesn't explicitly have it as a theorem. We didn't do GP and AP means in 3U, although there are exercises in cambridge on it. Is it required for 3U/4U?$

$\noindent The topic Harder 3U'' is actually a 4U topic (\underline{not} a 3U topic). As such, \color{blue}{the AM-GM inequality is \underline{only required for 4U} (not 3U)}\color{black}. (Inequalities are taught as a sub-topic in Harder 3U.) If you have a HSC 4U textbook, you should be able to find a chapter there on Harder 3U, and it should have the AM-GM inequality in it.$
• 10 Feb 2018, 2:07 PM
altSwift
Re: Complex Inequality
Quote:

Originally Posted by InteGrand
$\noindent The topic Harder 3U'' is actually a 4U topic (\underline{not} a 3U topic). As such, \color{blue}{the AM-GM inequality is \underline{only required for 4U} (not 3U)}\color{black}. (Inequalities are taught as a sub-topic in Harder 3U.) If you have a HSC 4U textbook, you should be able to find a chapter there on Harder 3U, and it should have the AM-GM inequality in it.$

Alright cool, thanks again!