Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
5  
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

3 Answers 3

up vote 4 down vote accepted

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

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

share|improve this answer
    
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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.