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# help

help me out by the what is minkowsky inequality

Note by Superman Son
3 years, 10 months ago

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The Minkowski's/Minkowsky's Inequality states that

$$(\displaystyle \sum_{k=1}^n |a_{k}+b_{k}|^{p})^{\frac {1}{p}} \leq (\displaystyle \sum_{k=1}^n |a_{k}|^{p})^{\frac {1}{p}}+( \displaystyle \sum_{k=1}^n |a_{k}|^{p})^{\frac {1}{p}}$$

So that we get $$\sqrt (a^{2}+(1-b)^{2}) + \sqrt (b^{2}+(1-c)^{2}) + \sqrt (c^{2}+(1-a)^{2} \geq \sqrt ((a+b+c)^2+((1-a)+(1-b)+(1-c))^2$$

= $$\sqrt ((a+b+c)^{2}+(3-(a+b+c))^{2}) = \sqrt (x^{2}+(3-x)^2)$$

= $$\sqrt (2(x- \frac {3}{2})^{2}+ \frac {9}{2}) \geq (\sqrt (\frac {9}{2})) = \frac {3\sqrt 2}{2}$$

where $$x=a+b+c$$.

Note: the square root is not completely overhead, but please excuse that. · 3 years, 10 months ago

but how u got the first line from minskowsky theorm · 3 years, 10 months ago

understood · 3 years, 10 months ago

thanks · 3 years, 10 months ago

as shourya showed you the proof of the inequality i'm only giving you link to understand what is minkowsky inequality...i think these could help you....but you yourself could search it in google...[!!!....:)]....this is better written in mathworld rather than wikipedia · 3 years, 10 months ago

Construct the following points on the plane: \begin{align} O&=(0,0)\\ X&=(a,1-b)\\ Y&=(a+b,2-b-c)\\ Z&=(a+b+c,3-a-b-c) \end{align} Now we have \begin{align} OX&=\sqrt{a^2+(1-b)^2}\\ XY&=\sqrt{b^2+(1-c)^2}\\ YZ&=\sqrt{c^2+(1-a)^2} \end{align} Also, $$Z$$ lies on the line $$y=3-x$$, and the minimal distance from $$O$$ to this line is the length of the perpendicular from $$O$$ to this line, which is $$\frac{3\sqrt2}{2}$$. Hence by the triangle inequality, \begin{align} \sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}&=OX+XY+YZ\\ &\geq OZ\\ &\geq\frac{3\sqrt2}{2}, \end{align} as desired. · 3 years, 10 months ago