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
–
Raja Metronetizen
·
3 years, 5 months ago

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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.
–
Ang Yan Sheng
·
3 years, 4 months ago

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@Ang Yan Sheng
–
Your method is elementary(and explanatory), and this same method is used to derive the Minkowski's Theorem.
–
Shourya Pandey
·
3 years, 4 months ago

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TopNewestThe 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. – Shourya Pandey · 3 years, 5 months ago

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– Superman Son · 3 years, 5 months ago

but how u got the first line from minskowsky theormLog in to reply

– Superman Son · 3 years, 5 months ago

understoodLog in to reply

– Superman Son · 3 years, 5 months ago

thanksLog in to reply

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 – Raja Metronetizen · 3 years, 5 months ago

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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. – Ang Yan Sheng · 3 years, 4 months ago

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– Shourya Pandey · 3 years, 4 months ago

Your method is elementary(and explanatory), and this same method is used to derive the Minkowski's Theorem.Log in to reply