- #1

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|α-β||γ|≤|α-γ||β|+|β-γ||α|,where |α|=√(α,α)

and point out when the equal mark holds.

Can someone help me out?

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- Thread starter tghg
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- #1

- 13

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|α-β||γ|≤|α-γ||β|+|β-γ||α|,where |α|=√(α,α)

and point out when the equal mark holds.

Can someone help me out?

- #2

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Abs(A-B)Abs(C) <= Abs(A)Abs(C) + Abs(C)Abs(B)

Abs(A-C)Abs(B) <= Abs(A)Abs(B) + Abs(C)Abs(B)

Abs(B-C)Abs(A) <= Abs(B)Abs(A) + Abs(C)Abs(A)

Abs(C)Abs(B) = Abs(A-C)Abs(B) - Abs(A)Abs(B)

Abs(A)Abs(C) = Abs(B-C)Abs(A) - Abs(B)Abs(A)

Abs(A-B)Abs(C) <= Abs(A-C)Abs(B) + Abs(B-C)Abs(A) - 2Abs(A)Abs(B)

I'm stumped about how to remove the -2Abs(A)Abs(B).

Equality if A=B=C

- #3

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Thanks very much!

Note that 2Abs(A)Abs(B)>=0, so

Abs(A-C)Abs(B) + Abs(B-C)Abs(A) - 2Abs(A)Abs(B) <= Abs(A-C)Abs(B) + Abs(B-C)Abs(A)

thus, Abs(A-B)Abs(C) <= Abs(A-C)Abs(B) + Abs(B-C)Abs(A) - 2Abs(A)Abs(B)

<= Abs(A-C)Abs(B) + Abs(B-C)Abs(A).

It's obvious that if A=B=C the equal mark holds.

But I think there are some other cases that satisfy the the equality.

Note that 2Abs(A)Abs(B)>=0, so

Abs(A-C)Abs(B) + Abs(B-C)Abs(A) - 2Abs(A)Abs(B) <= Abs(A-C)Abs(B) + Abs(B-C)Abs(A)

thus, Abs(A-B)Abs(C) <= Abs(A-C)Abs(B) + Abs(B-C)Abs(A) - 2Abs(A)Abs(B)

<= Abs(A-C)Abs(B) + Abs(B-C)Abs(A).

It's obvious that if A=B=C the equal mark holds.

But I think there are some other cases that satisfy the the equality.

Last edited:

- #4

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Abs(A-B)Abs(C) <= Abs(A-C)Abs(B) + Abs(B-C)Abs(A) - 2Abs(A)Abs(B)

to basic knowledge about numbers, e.g. 2Abs(A)Abs(B) >= 0, so

Abs(A-C)Abs(B) + Abs(B-C)Abs(A) - 2Abs(A)Abs(B) <= Abs(A-C)Abs(B) + Abs(B-C)Abs(A) which gives you what you want

Back to basics :p

- #5

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Abs(C)Abs(B) = Abs(A-C)Abs(B) - Abs(A)Abs(B)

Abs(A)Abs(C) = Abs(B-C)Abs(A) - Abs(B)Abs(A)

So, I'm very sorry to say that we didn't verify the inequality.

- #6

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The correct relationship is

Abs(C)Abs(B) >= Abs(A-C)Abs(B) - Abs(A)Abs(B)

Abs(A)Abs(C) >= Abs(B-C)Abs(A) - Abs(B)Abs(A)

but ASAICS this leads nowhere. :(

Abs(C)Abs(B) >= Abs(A-C)Abs(B) - Abs(A)Abs(B)

Abs(A)Abs(C) >= Abs(B-C)Abs(A) - Abs(B)Abs(A)

but ASAICS this leads nowhere. :(

Last edited:

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