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This question already has an answer here:

After what feels like an embarrassing hour of scribbling I can't seem to find a direct solution to the following problem

$Show \space that: a^2 + b^2 + c^2 \geq ab+bc+ca \space \space \forall [a,b,c] \in Z^{+}_0 $

I've tried placing each integer in an arbitrary order like so:

$a\leq b \therefore ab \leq b^2$

$b \leq c \therefore bc \leq c^2 \implies$

$a \leq c \therefore$ $ac \leq c^2$

Naturally I tried to add up the inequalities but this clearly yielded no results, but by nature of the third inequality I run into issues; what have I missed?

EDIT

I've constructed newer perhaps more insightful inequalities from one of the Dr's answer below:

$a-b \leq ab \leq b^2$

$b-c \leq bc \leq c^2$

$a-c \leq ac \leq c^2$

EDIT 2

While I've seen this flagged as a possible duplicate, this question appears towards the beginning of an intro book to mathematical proofs without any prior knowledge given, I feel aso though this should be provable using pure inequalities from first principals

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marked as duplicate by Martin R, Namaste, Harambe, Shailesh, Rolf Hoyer Nov 9 '17 at 4:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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HINT: this is equivalent to $$(a-b)^2+(b-c)^2+(c-a)^2\geq 0$$ which is true. this is also true for all real numbers $a,b,c$

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  • $\begingroup$ On second thought I've not found the answer $\endgroup$ – CertainlyNotAdrian Nov 8 '17 at 19:51
  • $\begingroup$ nice, send them to me or to this Forum for the other solvers $\endgroup$ – Dr. Sonnhard Graubner Nov 8 '17 at 19:53
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$$\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(2a^2-2ab)=\frac{1}{2}\sum_{cyc}(a^2-2ab+b^2)=\frac{1}{2}\sum_{cyc}(a-b)^2\geq0.$$

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  • $\begingroup$ While I'm inclined to accept Dr Sonnhard's answer, I'm curious as to what this is; can you please expand on this $\endgroup$ – CertainlyNotAdrian Nov 8 '17 at 19:50
  • $\begingroup$ @Adrian Yes of course. What is your question? $\endgroup$ – Michael Rozenberg Nov 8 '17 at 19:53
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    $\begingroup$ Duplicate answer See this same answer by the same user as we find here $\endgroup$ – Namaste Nov 8 '17 at 20:20
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    $\begingroup$ @MichaelRozenberg The right thing to have done here was to have identified the question as a duplicate. To answer a duplicate question you've already answered, with the identically same answer you posted on the original, simply indicates that you will do anything, even break site rules or consensus, to gain points. $\endgroup$ – Namaste Nov 8 '17 at 20:23
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    $\begingroup$ Not to mention that even the original is basically the same as the accepted answer. $\endgroup$ – Arnaud D. Nov 8 '17 at 20:31

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