Inequality

Hey guys,Im sitting in front of a task where I have sầu to prove that a²+b²+c²>=ab+bc+ca for a,b,c>=0. From what it looks like a direct proof seems like the way to lớn go, but so far Ive almost only done inductive proofs, so I need help because I have absolutely no idea. I dont even know a single transformation that would make sense.

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Note that your ineunique is equivalent to

a2+b2+c2-ab-ac-bc ≥ 0,

in turn equivalent to

(a2-2ab+b2) + (a2-2ac+c2) + (b2-2bc+c2) ≥ 0.

It is only a couple of more steps lớn prove sầu the desired ineunique.


a2+b2+c2-ab-ac-bc ≥ 0,

Didnt know you could transform it that easily with inequalities, Ill start from that point! Thanks!


So... from your second equation I easily get lớn the solution:

(a-b)²+(a-c)²+(b-c)²>=0

All three parts of the sum are obviously larger than or equal to lớn 0.However, I dont underst& how you got from

a²+b²+c²-ab-ac-bc ≥ 0,

there.These require 2a²+2b²+2c² khổng lồ begin with.


You can assume that a >= b >= c. If this is false, switch some numbers around until it's true.

I'm sure there's a more elegant solution, but you can:

Prove sầu the case where a = b = c

Prove sầu the case where a > b = c

Prove the case where a = b > c

Prove the case where a > b > c


You can assume that a >= b >= c. If this is false, switch some numbers around until it's true.

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Figured that.

Prove sầu the case were a=b=c

Easy, for obvious reasons

2. Prove sầu the case where a>b=c

brings me khổng lồ a²+b²>=ab+abSo my problem here would be making a general solution. Id now just put a=b+1(b+1)²+b²=2b²+2b+1>=2b²+2b, but that would not be a general answer.

3. Prove the case were a=b>c

Same, would result in a²+c²>=ac+ac

4. Prove the case were a>b>c

No idea.

I guess the problem is that I just dont know how to work with the ">" operator for a proof.


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