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Proving that for reals $a,b,c$, $(a+b+c)^2\leq 3(a^2+b^2+c^2)$.

This is a homework question and I have no clue where to even start on this. I don't know if I am just tired or what but I can't get anywhere. I've expanding both sides and seeing if that gets me anywhere but I don't see how it helps.

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Hint: Expand (a good start!). Move all the stuff to the r.h.s. Recall that $0\le (x-y)^2=x^2-2xy+y^2$. Identify some terms... – Jyrki Lahtonen Sep 19 '11 at 4:38
up vote 4 down vote accepted

Hint: $$0\le(a-b)^2+(b-c)^2+(c-a)^2$$

$$\implies0\le2(a^2+b^2+c^2)-2(ab+bc+ca)\qquad\text{(foil)}$$ $$\implies a^2+b^2+c^2+2(ab+bc+ca)\le3(a^2+b^2+c^2)\qquad\text{(rearrange, add squares)}$$ $$\implies (a+b+c)^2\le3(a^2+b^2+c^2)\qquad\text{(factor)}$$

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Thanks! I got it with this and the comment in the OP :) – Logan Serman Sep 19 '11 at 4:52

One way: $(a + b + c)$ is the dot product of $(a,b,c)$ and $(1,1,1)$, and you have $|v \cdot w| = ||v||\, ||w||\,|\cos(\theta)| \leq ||v|| \,||w||$ for any vectors $v$ and $w$, where $\theta$ is the angle between the two vectors. This is a form of the Cauchy-Schwarz inequality.

You can also just expand it and use the arithmetic-geometric mean inequality in the right way.

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We need to expand both sides of inequality:

$(a+b)^2+c^2+2(a+b)c\leq 3a^2+3b^2+3c^2$

$a^2+b^2+c^2+2ab+2ac+2bc\leq 3a^2+3b^2+3c^2$

$2ab+2ac+2bc\leq 2a^2+2b^2+2c^2$

$0\leq a^2+b^2-2ab+a^2+c^2-2ac+b^2+c^2-2bc$

$0\leq (a-b)^2 +(a-c)^2 +(b-c)^2$ , what is obvious true.

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