33But still if theres square in RHS then why thats | |
:P
Prove this tough looking and sounding Cauchy–Schwarz inequality:
If x1,...,xn ε R and y1,...,yn ε R
|x1y1+...+xnyn|2 ≤ (|x1|2+...+|xn|2) (|y1|2+....+|yn|2)
Comment on the equality!
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8 Answers
62x's and y's are all reals..
There is a more general version.. but we are preparing for IIT JEE not INMO :)
33If x1,...,xn ε R and y1,...,yn ε R are the components of x and y ;
samajh me nahi aaya :( " components"
mera dimag aaj anhi chal raha lagta hai :(
62arrey wo componenets etc was mistyping.. or u could say it was a more general result.. i have removed that "crap"
39{x12+x22+...................+xn2}{y12+y22+.............+yn2} - {x1y1+.......................xnyn}2
=1/2 Σ{xp2yq2+xq2yp2-2xpypxqyq} note that inside Σ p≠q and p,q=1,2,3,4,.......................n
the equality sign holds iff xpyq-xqyp=0 i.e. iff xp:xq::yp:yq , for every pair of suffixes p,q i.e.e iff x1:x2:.............:xn::y1:y2:...............yn
this is the first of the proofs i have