1cauchy scwrtz 'equality' holds wen a=b=c, 1=w=w2...
which is not possible!
find the min value of \left|a+b\omega +c\omega ^{2} \right|
where a,b,c are not equal integers and ω≠1 is cube root of unity..
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31 Answers
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1Thank you Sri
that gets me to understand the soln[1][1]
But still..........What does
"since minimum diffrnce betwn two terms is 1" mean and where does it come from in wats in name's post[7][7]
1\left| a+b\omega +c\omega^{2}\right| = √(a-b/2-c/2)2 + 3(c-b)2/4=√((a-b)2 + (b-c)2 +( c-a)2)/2
the expression is min. when 2 of the a,b,c are equal
let a=b and b-c=±1 & c-a=±1
so the answer is 1
1let the iven exprsn be = Z
i.e \left|Z \right| = x
thus x^{2}= Z\bar{Z}
wihch reduces to the above one becos the conjugate ofa+b\omega +c\omega ^{2} is a+b\omega ^{2}+c\omega
so this solves it na ???
1it depends on this one It think
a^{2}+b^{2}+c^{2}-ab-bc-ca = (a+b\omega +c\omega ^{2})(a+ b\omega ^{2}+c\omega )
that's all wat's in the name has used.
1nameless are U kidding or is that a serious answer
Can U explain it if it actually is ......[1]
1x=mod of that crap|:P
x^2=square of dat crap :P
so nw
itz
x^2=1/2[(a-b)^2+(b-c)^2+(c-a)^2]
nw a b c r nt equal simulatneously
so d value of dat xperssion is > or = (1+1+0)
so itz > or =1
since minimum diffrnce betwn two terms is 1
so answer is 1
:P
1here though a,b,c are variables but 1,w,w2 are not...
and also given that a≠b≠c ...
we are lucky enuf to get 0 on LHS by properties of complex no..
but that doesnt mean we can use equality here..
1i dnt noe this thm had done it longggggggggggggggggggg back..........
bt iit mein ni ahi so ditch mara :P
1chauchy - shtwartzur ytu .. some inequality with this kinda name
(a2 + b2 + c2 )(12 + ω2 + (ω2)2) ≥ (a + bω + cω2)2
= 0 ≥ (a + bω + cω2)2
but |a + bω + cω2| not -ve so it is = 0