Analysis Help; proofs via axioms

[silvermane]

Analysis Help; proofs via axioms :)

1. The problem statement:

Prove that for any real numbers a, b, c,

(a+b+c)^2\leq3*(a^2 +b^2+c^2)​

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
0\leq(a + b)^2

So can we say that
0\leq(a + b + c)^2
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

(a_{1}+a_{2}+a_{3})^2\leq3*(a_{1}^2 +a_{2}^2+a_{3}^2)​

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!

 

[VeeEight]

A good way to work on a problem like this is to simplify it the only ways you know how and then work backwards, maybe modifying your work, to justify your steps. Expand the left hand side. You will get three squares and a multiple of 2. Now use your property (3) to cancel out the three squares and consider property (4). I hope things have cleared up after this. After you've seen the way you can work backwards to justify your steps.

 

[silvermane]

A good way to work on a problem like this is to simplify it the only ways you know how and then work backwards, maybe modifying your work, to justify your steps. Expand the left hand side. You will get three squares and a multiple of 2. Now use your property (3) to cancel out the three squares and consider property (4). I hope things have cleared up after this. After you've seen the way you can work backwards to justify your steps.

Haha yes, this is more than helpful. Thank you for this hint, it will be well used, and is well appreciated.

 

[VeeEight]

Great! Glad you have worked it out