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

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.

## Comments

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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.

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but how u got the first line from minskowsky theorm

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understood

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thanks

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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

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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.

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

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