• tghg

tghg

If α,β,γ are vectors in the Euclid space V, please show that
|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?

This question reeks of Triangle Inequality, I'm assuming the relation: Abs(A-B) <= Abs(A) + Abs(B)

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

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.

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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

Oh,my god!I found a severe mistake. We can't claim that
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.

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. :(

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