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.

Prove that for any real number $a$, $b$, $c$, $x$, $y$ and $z$, there results $(ax + by + cz)^2 \leq (a^2 + b^2 + c^2)(x^2 + y^2 + z^2)$

I have thought this Q for long time but I still can't get the answer. Can anyone help me please? Thank you!~

share|improve this question
4  
See Cauchy-Schwarz inequality –  Frank Science Jul 5 '12 at 13:40
    
Another proof is elementary: consider $f(t)=(a^2+b^2+c^2)t^2-2(ax+by+cz)t+(x^2+y^2+z^2)=(at-x)^2+(bt-y)^2+(ct-z)^2\ge0‌​$ whenever $t\in\Bbb R$. –  Frank Science Jul 5 '12 at 13:41
    
Another proof is about Lagrange's identity. –  Frank Science Jul 5 '12 at 13:43
2  
@FrankScience: that's basically the same idea as the proof of Cauchy-Schwarz, isn't it? –  Ben Millwood Jul 5 '12 at 13:47

3 Answers 3

up vote 0 down vote accepted

We observe that \begin{eqnarray*} \text{RHS}&=&(a^2x^2+b^2y^2+c^2z^2)+(a^2y^2+b^2x^2)+(a^2z^2+c^2x^2)+(+b^2z^2+c^2y^2)\\ &=&(a^2x^2+b^2y^2+c^2z^2+2axby+2bycz+2czax)+(a^2y^2+b^2x^2-2aybx)+(a^2z^2+c^2x^2-2azcx)+(b^2z^2+c^2y^2-2bzcy)\\ &=&(ax+by+cz)^2+(ay-bx)^2+(az-cx)^2+(bz-cy)^2\\ &\geq&(ax+by+cz)^2\\ &\geq&\text{LHS}. \end{eqnarray*}

share|improve this answer

This question is direct application of Cauchy-Schwarz inequality.

share|improve this answer

Let $u=(a,b,c)$ and $v=(x,y,z)$ be two vectors in $\mathbb{R}^3$. We have $$ (\left<u,v\right>)^2=(ax+by+cz)^2 \text{ and } \|u\|^2\|v\|^2=(a^2+b^2+c^2)(x^2+y^2+z^2). $$ Since $(\left<u,v\right>)^2\leq \|u\|^2\|v\|^2$ we have the given inequality.

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.