Can someone prove this?

  • Thread starter totoro
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  • #1
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my teacher gave me a question in the class...

(1/a)+(1/b)+(1/c)=1/(a+b+c)
then prove:
(1/a^2002)+(1/b^2002)+(1/c^2002)=1/(a^2002 + b^2002 + c^2002)

can someone help me to prove it?


kian yew.
 

Answers and Replies

  • #2
HallsofIvy
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What conditions are put on a, b, c?


It's fairly easy to prove that (1/a)+(1/b)+(1/c)=1/(a+b+c)
is never true for a, b, c positive and that, therefore,

(1/a^2002)+(1/b^2002)+(1/c^2002)=1/(a^2002 + b^2002 + c^2002)

is never true for a, b, c real.
 
  • #3
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she just gave me this. why it's never true for a,b,c is real?
can't a,b,c be any real numbers
 
  • #4
Njorl
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In general 1/a+1/b+1/c=1/(a+b+c) is not generally true. I am assuming that it is supposed to be a given that it is true for this problem. Is this so?

If so, b=a and c=-a is a workable solution set for the first assumption. However, it is a counter-example for the second equation. The second equation is disproved, not proved.

Njorl
 
  • #5
Njorl
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Actually, the general solution for 1/a+1/b+1/c=1/(a+b+c) is:

a=-b or a=-c or b=-c with no constraint on the other variable.

If the second equation used odd powers instead of even powers, it would be true, given the first equation.

Njorl
 
  • #6
44
1
Originally posted by Njorl
If so, b=a and c=-a is a workable solution set for the first assumption. However, it is a counter-example for the second equation. The second equation is disproved, not proved.
Take a+b=0 and let c free. Then the second is satisfied because its trivial. These are indeed the only possible solution types which can be applied to this.
 
  • #7
Njorl
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The second equation is not satisfied, it is disproved. if a+b=0, a=-b.

Then, when each is raised to an even power like 2002, the sign change disappears. An odd power would make it work.

Njorl
 
  • #8
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Oh ****! I read 1/(x+y+z)^2002

You're right. Momentaneous offuscation :frown:
 
  • #9
HallsofIvy
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(1/a)+(1/b)+(1/c)= (ab+ ac+ bc)/abc so
if (1/a)+(1/b)+(1/c)=1/(a+b+c) we have
(ab+ac+bc)/abc= 1/(a+b+c)

Now multiply both sides of the equation by abc and a+b+c to "clear the fractions":

(a+b+c)(ab+ac+bc)= abc
Multiply the left side out:
a2b+ a2c+ abc+ ab2+abc+b2c+ abc+ ac2+ bc2= abc.

One of those 3 "abc"s on the left cancels the one on the right so
ab2+b2c+ ac2+ bc2+ 2abc= 0. If a, b, c are all real, this is impossible.

In particular, in (1/a^2002)+(1/b^2002)+(1/c^2002)=1/(a^2002 + b^2002 + c^2002) all the denominators are to even powers and so are positive as long as a, b, are real.
 
  • #10
climbhi
Originally posted by HallsofIvy
[B so
ab2+b2c+ ac2+ bc2+ 2abc= 0. If a, b, c are all real, this is impossible. [/B]
Okay I followed you to here, but can't see why that is impossible for all reals. Maybe I'm missing something, could you help me out.
 
  • #11
FZ+
1,561
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Indeed, that seems to me to be wrong. (Though I might just be mistaken). Can't you rewrite that as 0 = (b + c)(ab+ac+bc) ?
Then clearly if b+c = 0, the equation is correct...
 
  • #12
dg
(1/a)+(1/b)+(1/c)=1/(a+b+c)

requires a, b, and c and a+b+c to be non-zero for the divisions to be defined. Said that, we can proceed by passing to a common denominator on the left side

(ab+ac+bc)/abc= 1/(a+b+c)

Now multiply both sides of the equation by abc and a+b+c to "clear the fractions" which is possible only in presence of the previous conditions!!!

(a+b+c)(ab+ac+bc)= abc

Multiply out the left side

a^2b+a^2c+abc+ab^2+abc+b^2c+abc+ac^2+bc^2= abc.

simplify

a^2b+a^2c+ab^2+2abc+b^2c+ac^2+bc^2+2abc= 0

collect c from all the terms where it appears with power 1

a^2b+ab^2+ac^2+bc^2+(a^2+2ab+b^2)c= 0

recognize square of binomial and collect c^2 terms and ab from the two remaining ones

ab(a+b)+(a+b)c^2+(a+b)^2c= 0

collect (a+b)

(a+b)[ c^2+(a+b)c+ab ]= 0

It easy to see if you that the second term can be rewritten as a product of two binomials and have

(a+b)(b+c)(c+a)= 0

That requires that at least one (but not all three of them, otherwise all should be zeros against our hypotheses) of the following holds

a=-b aut b=-c aut c=-a

For(1/a^2002)+(1/b^2002)+(1/c^2002)=1/(a^2002 + b^2002 + c^2002) all you need to do is to substitute a->a^2002, etc. in this case by the way the previous formulas cannot hold for any number different from zero and hence the equality is always impossible.

Hope this helped
 
  • #13
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thanks to all your help, but i'm thinking another way to do it and i don't know how to become this:

first, turn (1/a)+(1/b)+(1/c)=1/(a+b+c) into
(1/a^2)+(1/b^2)+(1/c^2)=1/(a^2+b^2+c^2) then turn again into (1/a^3)+(1/b^3)+(1/c^3)=1/(a^3+b^3+c^3) and then.....
from this i can say that (1/a^n)+(1/b^n)+(1/c^n)=1/(a^n+b^n+c^n). the problem for me is i don't know how to make the first few line that i show above. or this way is wrong? can i use induction method?


kian yew.
 
  • #14
dg
Originally posted by totoro
thanks to all your help, but i'm thinking another way to do it and i don't know how to become this:

first, turn (1/a)+(1/b)+(1/c)=1/(a+b+c) into
(1/a^2)+(1/b^2)+(1/c^2)=1/(a^2+b^2+c^2) then turn again into (1/a^3)+(1/b^3)+(1/c^3)=1/(a^3+b^3+c^3) and then.....
from this i can say that (1/a^n)+(1/b^n)+(1/c^n)=1/(a^n+b^n+c^n). the problem for me is i don't know how to make the first few line that i show above. or this way is wrong? can i use induction method?


kian yew.
Yes it is wrong!
please check my previous answer!

induction here is a useless complication all you need to consider is that is an even power

Dario
 
  • #15
152
0
Originally posted by totoro
thanks to all your help, but i'm thinking another way to do it and i don't know how to become this:

first, turn (1/a)+(1/b)+(1/c)=1/(a+b+c) into
(1/a^2)+(1/b^2)+(1/c^2)=1/(a^2+b^2+c^2) then turn again into (1/a^3)+(1/b^3)+(1/c^3)=1/(a^3+b^3+c^3) and then.....
from this i can say that (1/a^n)+(1/b^n)+(1/c^n)=1/(a^n+b^n+c^n). the problem for me is i don't know how to make the first few line that i show above. or this way is wrong? can i use induction method?


kian yew.
you wrong
i think this cant use induction
because when n=1, this equation also cant do
so i think maybe a=b=c=1 .....:)
 
  • #16
499
1
I guess it doesn't work for anything.
 
Last edited:
  • #17
152
0
sorry

sorry....
mistake...a,b,c cant be 1....
because the 1/1 + 1/1 + 1/1 cant be 1/(1+1+1)
haha....so i think this equation are no solve......
 

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