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Math Help - Triangle inequality help

  1. #1
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    Triangle inequality help

    Hi following a problem on a past exam and have come across a step i can't follow with the triangle inequality,

    | 5x - y + (xy.cos(y))/(x^1/2 + y^1/2) |

    is

    <= 5|x| + |y| + |x|.|y|.|cos(y)|/(x^1/2 + y ^1/2)
    and says it's via the trianlge inequality if someone can explain why it would be great cheers.
    Last edited by monster; June 22nd 2011 at 01:21 AM.
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  2. #2
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    Re: Triangle inequality help

    Quote Originally Posted by monster View Post
    Hi following a problem on a past exam and have come across a step i can't follow with the triangle inequality,

    | 5x - y + (xy.cos(y))/(x^1/2 + y^1/2) |

    is

    <= 5|x| + |y| + |x|.|y|.|cos(y)|/(x^1/2 + y ^1/2)
    and says it's via the trianlge inequality if someone can explain why it would be great cheers.
    Dear monster,

    The triangular inequality says for any two real numbers a,b;

    \mid a+b\mid\leq\mid a\mid+\mid b\mid

    Applying the triangular inequality, replacing a\rightarrow{5x-y}\mbox{ and }b\rightarrow \frac{xy\cos(y)}{\sqrt{x}+\sqrt{y}}

    \left|5x - y + \frac{xy\cos(y)}{\sqrt{x}+\sqrt{y}}\right|\leq \left|5x-y\right|+\left|\frac{xy\cos(y)}{\sqrt{x}+\sqrt{y}}  \right|

    Applying the triangular inequality again for \mid 5x-y\mid;

    \left|5x - y + \frac{xy\cos(y)}{\sqrt{x}+\sqrt{y}}\right|\leq \left|5x-y\right|+\left|\frac{xy\cos(y)}{\sqrt{x}+\sqrt{y}}  \right|\leq 5\mid x\mid+\mid  y\mid+\left|\frac{xy\cos(y)}{\sqrt{x}+\sqrt{y}} \right|

    \mid\sqrt{x}+\sqrt{y}\mid=\sqrt{x}+\sqrt{y}\mbox{ since }\sqrt x>0\mbox{ and }\sqrt y>0~\forall~x,y\in \Re^+

    Therefore, \left|5x - y + \frac{xy\cos(y)}{\sqrt{x}+\sqrt{y}}\right|\leq 5\mid x\mid+\mid  y\mid+\frac{\mid x\mid\mid y\mid\mid\cos(y)\mid}{\sqrt{x}+\sqrt{y}}
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  3. #3
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    Re: Triangle inequality help

    yes, but when you re-apply for |5x - y| how does it work? it seems what you have written would work for |5x + y| but what about with the negative in there??
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  4. #4
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    Re: Triangle inequality help

    Quote Originally Posted by monster View Post
    yes, but when you re-apply for |5x - y| how does it work? it seems what you have written would work for |5x + y| but what about with the negative in there??
    In this case your a is 5x and your b is -y.

    \mid 5x-y\mid=\mid 5x+(-y)\mid \leq\mid 5x\mid+\mid -y\mid\leq 5\mid x\mid+\mid y\mid~;\mbox{ since }\mid-y\mid=\mid y\mid
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