<s>[INST] 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. 
 [/INST] It follows immediately from Cauchy-Schwarz:
$$(a+b+c)^2 = (1\cdot a + 1 \cdot b + 1 \cdot c)^2\leq (1^2+1^2+1^2) (a^2 + b^2 + c^2)$$
</s>