1On simplification the expression
(ap-b)2+(bp-c)2+(cp-d)2≤0
results:-
shouldn't this expression always be ≥0 ??
Could this problem be correct?
if a,b,c,d and p are different real numbers such that
(a2+ b2+c2)p2 -2(ab+bc+cd) p +(b2+c2+d2) ≤ 0 then
a,b,c and d are in geometric progession
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11 Answers
11On simplification the expression
(ap-b)2+(bp-c)2+(cp-d)2≤0
results:-
shouldn't this expression always be ≥0 ??
62yes precisely.
Now when u say x≥0 and x≤0
what is the only option for this to be true!
isnt it x=0!
so we have (ap-b)2+(bp-c)2+(cp-d)2=0
and for this
(ap-b)2=0 and (bp-c)2=0 and (cp-d)2=0
thus ap=b, bp=c and cp=d
1Thanks!
But the expression can never be less than 0 , so isn't that part incorrect?
62hmm...
see, it is not that the expression {say f(x)} is less than 0...
It is like we know that f(x) ≤0 ... but then we also know that f(x) is not less than 0... so the only way this is possible (or the only possible thing that will not lead to a contradiction) is that x=0...
I hope this makes some more sense.. if not ask me again :)
I will try to find some mroe way to explain :)
1Yes that does make sense, but I still feel the ≤ part should have been ≥, otherwise the mathematical condition given in the problem is incorrect.
Anyway thanks a lot I got the solution, through TargetIIT. My tutor too was stuck.
Will be back with more
62i think u should understand this...
I will give an example...
if i said (x-a)2≤0, is it wrong? what will this imply?
Does this not imply the only way this could be true is x=a ?
other wise it will be a contradiction....
(See u have a point when u say that it should be ≥0)
But if that were the question, the whole thing would collapse!!
Try to solve this same problem if we take the case that you are talking... Then will u be able to make any such conclusion...??