Proving the following quadratic inequality? Apologies if this has been asked before - I could not find a question with this exact inequality.
Basically the inequality is
$$(a+b+c)^2 \leq 3 a^2 + 3 b^2 + 3 c^2$$
Expanding it out we see that
$$(a+b+c)^2 = a^2 +b^2 + c^2 + 2ab + 2bc + 2ac$$
so I guess it is equivalent to showing that
$$ab + bc + ac \leq a^2 + b^2 + c^2$$
Which makes sense to me. But how exactly do I prove it?
We can assume WLOG that each $a,b,c > 0$ since $ab \leq |a||b|$. From here, I guess we need to show that
$$ab \leq \frac{1}{2} \left(\max(a,b)^2 + \min(a,b)^2
\right)$$
And the result follows by adding up each term. But I'm not really sure why this must hold. 

Your last step follows easily from $(a-b)^{2} \geq 0$. (Consider the cases $a<b$ and $a \geq b$).
