- #1

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- 0

(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.

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- Thread starter totoro
- Start date

In summary, the conversation discusses a mathematical proof involving the equations (1/a)+(1/b)+(1/c)=1/(a+b+c) and (1/a^2002)+(1/b^2002)+(1/c^2002)=1/(a^2002 + b^2002 + c^2002). The experts in the conversation explain why the second equation is disproved, rather than proved, and provide counter-examples to support their explanation. One person also suggests a different approach to solve the problem, but it is deemed incorrect. Induction is deemed unnecessary for the proof.

- #1

- 42

- 0

(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.

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- #2

Science Advisor

Homework Helper

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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

- 42

- 0

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

can't a,b,c be any real numbers

- #4

Science Advisor

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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

Science Advisor

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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

Science Advisor

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

Njorl

- #8

- 44

- 1

Oh ****! I read 1/(x+y+z)^2002

You're right. Momentaneous offuscation

You're right. Momentaneous offuscation

- #9

Science Advisor

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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:

a

One of those 3 "abc"s on the left cancels the one on the right so

ab

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

Originally posted by HallsofIvy

[B so

ab^{2}+b^{2}c+ ac^{2}+ bc^{2}+ 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

- 1,600

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Then clearly if b+c = 0, the equation is correct...

- #12

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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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

Originally posted by totoro

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

- #15

- 152

- 0

Originally posted by totoro

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 can't use induction

because when n=1, this equation also can't do

so i think maybe a=b=c=1 ...:)

- #16

- 502

- 1

I guess it doesn't work for anything.

Last edited:

- #17

- 152

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sorry...

mistake...a,b,c can't be 1...

because the 1/1 + 1/1 + 1/1 can't be 1/(1+1+1)

haha...so i think this equation are no solve...

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