Complex Inequality (1 Viewer)

altSwift · Feb 10, 2018

Confused with iii). Any help would be appreciated!

InteGrand · Feb 10, 2018

altSwift said:
View 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!)$

altSwift · Feb 10, 2018

InteGrand said:
$\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

altSwift · Feb 10, 2018

InteGrand said:
$\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$$

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

InteGrand · Feb 10, 2018

altSwift said:
$I mangaed to get this after lots of tries$

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

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

$$ 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 $x +\frac{1}{x}\geq 2$.$

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

altSwift · Feb 10, 2018

InteGrand said:
$\noindent Well if we want to show that$

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

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

$$$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.$

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

InteGrand · Feb 10, 2018

altSwift said:
$\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.$

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

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

altSwift · Feb 10, 2018

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

Alright cool, thanks again!

Complex Inequality (1 Viewer)

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