# Proving a seemingly simple inequality is proving difficult [duplicate]

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

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

• Apply AM/GM to $a^2$ and $b^2$, and then to the other couple of similar pairs. – Lord Shark the Unknown Nov 8 '17 at 19:41
• @LordSharktheUnknown AM/GM? – CertainlyNotAdrian Nov 8 '17 at 19:41
• Arithmetic mean geometric mean inequality – Zach Boyd Nov 8 '17 at 19:52
• a possible duplicate or a real one? – Dr. Sonnhard Graubner Nov 8 '17 at 19:55
• – Martin R Nov 8 '17 at 19:56

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