Can someone prove this?

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.

HallsofIvy
Homework Helper
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.

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

Njorl
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

Njorl
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

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.

Njorl
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

You're right. Momentaneous offuscation

HallsofIvy
Homework Helper
(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.

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.

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...

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

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.

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!

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

Dario

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 .....:)

I guess it doesn't work for anything.

Last edited:
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......