# Proving the following quadratic inequality? [duplicate]

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.

## marked as duplicate by Martin R, Cyclohexanol., stressed out, user10354138, Joel Reyes NocheDec 19 '18 at 13:41

Your last step follows easily from $$(a-b)^{2} \geq 0$$. (Consider the cases $$a and $$a \geq b$$).
It follows immediately from Cauchy-Schwarz: $$(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)$$
Once you get your final inequality you simply multiply both sides by $$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}.$$
• And in fact multiplying by $2$ is only necessary because of having divided by $2$ previously! I went straight through $2ab+2bc+2ac\le 2a^2+2b^2+2c^2$ when I attempted it, and totally missed the massive clue in $-2ab$. I saw all those $2$'s and instinctively divided. – timtfj Dec 19 '18 at 16:46
• yes I was going to mention why he divided the $2$ but then I just decided to leave it – Prakhar Nagpal Dec 19 '18 at 18:30
It is $$(a-b)^2+(b-c)^2+(c-a)^2\geq 0$$