Proving the following quadratic inequality?

Question

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.

Yes we will prove $a^2+b^2+c^2\ge ab+bc+ca$ and it is equivalent to $(a-b)^2+(b-c)^2+(c-a)^2\ge 0$ — Word Shallow, Dec 19 '18 at 08:22
A more general case including your one can be found here: https://math.stackexchange.com/questions/1249986/prove-qm-am-inequality — trancelocation, Dec 19 '18 at 08:25
Another one: Proving that for reals $a,b,c$, $(a + b + c)^2 \leq 3(a^2+b^2+c^2)$ — Martin R, Dec 19 '18 at 08:33

score 7 · Answer 1 · answered Dec 19 '18 at 08:29

7

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

answered Dec 19 '18 at 08:29

trancelocation

32,243

score 6 · Answer 2 · answered Dec 19 '18 at 08:21

6

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

answered Dec 19 '18 at 08:21

Kavi Rama Murthy

311,013

score 4 · Answer 3 · answered Dec 19 '18 at 08:27

4

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 $\forall \; a,b,c \; \epsilon \; \mathbb{R}.$

answered Dec 19 '18 at 08:27

Prakhar Nagpal

1,312

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

score 2 · Answer 4 · answered Dec 19 '18 at 08:22

2

It is $$(a-b)^2+(b-c)^2+(c-a)^2\geq 0$$

answered Dec 19 '18 at 08:22

Dr. Sonnhard Graubner

95,283

Proving the following quadratic inequality?

4 Answers4