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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
Construct the following points on the plane:
OXYZ=(0,0)=(a,1−b)=(a+b,2−b−c)=(a+b+c,3−a−b−c)
Now we have
OXXYYZ=a2+(1−b)2=b2+(1−c)2=c2+(1−a)2
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 232. Hence by the triangle inequality,
a2+(1−b)2+b2+(1−c)2+c2+(1−a)2=OX+XY+YZ≥OZ≥232,
as desired.
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The Minkowski's/Minkowsky's Inequality states that
(k=1∑n∣ak+bk∣p)p1≤(k=1∑n∣ak∣p)p1+(k=1∑n∣ak∣p)p1
So that we get (a2+(1−b)2)+(b2+(1−c)2)+(c2+(1−a)2≥((a+b+c)2+((1−a)+(1−b)+(1−c))2
= ((a+b+c)2+(3−(a+b+c))2)=(x2+(3−x)2)
= (2(x−23)2+29)≥((29))=232
where x=a+b+c.
Note: the square root is not completely overhead, but please excuse that.
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thanks
but how u got the first line from minskowsky theorm
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understood
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: OXYZ=(0,0)=(a,1−b)=(a+b,2−b−c)=(a+b+c,3−a−b−c) Now we have OXXYYZ=a2+(1−b)2=b2+(1−c)2=c2+(1−a)2 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 232. Hence by the triangle inequality, a2+(1−b)2+b2+(1−c)2+c2+(1−a)2=OX+XY+YZ≥OZ≥232, as desired.
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Your method is elementary(and explanatory), and this same method is used to derive the Minkowski's Theorem.