Q:

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.  

A:

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

A:

It follows immediately from Cauchy-Schwarz:
$$(a+b+c)^2 = (1\cdot a + 1 \cdot b + 1 \cdot c)^2\leq (1^2+1^2+1^2) (a^2 + b^2 + c^2)$$

A:

Once you get your final inequality you simply multiply both sides by $2$ and you get $$2a^2 + 2b^2 + 2c^2 - 2ab -2bc-2ca \geq 0$$ This can be rewritten as $$a^2 + b^2 - 2ab + a^2 + c^2 - 2ac + b^2 + c^2 -2ca\geq 0$$ Which translates to $$\Biggl(\Bigl(a-b\bigl)^2 + \Bigl(b-c\Bigl)^2 + \Bigl(c-a\bigl)^2\Biggl) \geq 0$$ Which is true $\forall \; a,b,c \; \epsilon \; \mathbb{R}.$ 

