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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

CertainlyNotAdrian
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2 Answers2

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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 $a,b,c$

Dr. Sonnhard Graubner
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$$\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(2a^2-2ab)=\frac{1}{2}\sum_{cyc}(a^2-2ab+b^2)=\frac{1}{2}\sum_{cyc}(a-b)^2\geq0.$$

Michael Rozenberg
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  • While I'm inclined to accept Dr Sonnhard's answer, I'm curious as to what this is; can you please expand on this – CertainlyNotAdrian Nov 08 '17 at 19:50
  • @Adrian Yes of course. What is your question? – Michael Rozenberg Nov 08 '17 at 19:53
  • @MichaelRozenberg Firstly what are the sums referring to? – CertainlyNotAdrian Nov 08 '17 at 20:05
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    Duplicate answer See this same answer by the same user as we find here – amWhy Nov 08 '17 at 20:20
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    @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. – amWhy Nov 08 '17 at 20:23
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    Not to mention that even the original is basically the same as the accepted answer. – Arnaud D. Nov 08 '17 at 20:31
  • @Adrian It's a cyclic sum. For example, for three variables we obtain: $\sum\limits_{cyc}a=a+b+c$; $\sum\limits_{cyc}ab=ab+bc+ca$; $\sum\limits_{cyc}(a^2-ab)=a^2-ab+b^2-bc+c^2-ca$. Good luck! – Michael Rozenberg Nov 09 '17 at 03:51