- #1
silvermane
Gold Member
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Analysis Help; proofs via axioms :)
1. The problem statement:
Prove that for any real numbers a, b, c,
2. These are the axioms we are permitted to use:
01) Exactly one of these hold: a<b, a=b, or b<a
02) If a<b, and b<c, then a<c
03) If a<b, then a+c < b+c for every c
04) If a<b and 0<c, then ac<bc.
It follows from problem number 1, that
[tex]0\leq(a + b)^2[/tex]
So can we say that
[tex]0\leq(a + b + c)^2[/tex]
and would this help me to solve the proof? I'm having a great deal of trouble getting used to thinking like this; I'm very used to thinking combinatorially.
I do know that this is a special case of the Cauchy-Schwartz Sequence where we have
Any help/hints would be greatly appreciated, and I always do my best to return the favor! Thank you so much for your time!
1. The problem statement:
Prove that for any real numbers a, b, c,
[tex](a+b+c)^2\leq3*(a^2 +b^2+c^2)[/tex]
2. These are the axioms we are permitted to use:
01) Exactly one of these hold: a<b, a=b, or b<a
02) If a<b, and b<c, then a<c
03) If a<b, then a+c < b+c for every c
04) If a<b and 0<c, then ac<bc.
The Attempt at a Solution
It follows from problem number 1, that
[tex]0\leq(a + b)^2[/tex]
So can we say that
[tex]0\leq(a + b + c)^2[/tex]
and would this help me to solve the proof? I'm having a great deal of trouble getting used to thinking like this; I'm very used to thinking combinatorially.
I do know that this is a special case of the Cauchy-Schwartz Sequence where we have
[tex](a_{1}+a_{2}+a_{3})^2\leq3*(a_{1}^2 +a_{2}^2+a_{3}^2)[/tex]
I'm just not allowed to use that fact to proove it Any help/hints would be greatly appreciated, and I always do my best to return the favor! Thank you so much for your time!