2005.......complex

Thank 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]

\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

let 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 ???

it 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.

x=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

processing........... # # #

anyone has a solution ..

so what to do ..

cauchy scwarz inequality holds good for all reals......

ok sky go on !!

here 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..

can u point out the error wats i n ur name ?

eureka aren't u intrested !!

yaar then how come i got a contradiction !

chauchy - 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

i got 0 ..

asnwer is...1...........

memory test :P

okzzzz