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how can one show that

$(\bf{u}\cdot\bf{v})^2 \leq |\bf{u}|^2 |\bf{v}|^2$ and hence $|\bf{u}\cdot\bf{v}| \leq |\bf{u}| |\bf{v}|$

by using the inequality $|\bf{u}|^2 + 2x(\bf{u}\cdot\bf{v}) +x^2|\bf{v}|^2\geq 0$.

To use this inequality i tried to show that if $a+bx+cx^2 \geq 0$ for any real number x then $b^2 - 4ac \leq 0$ and to show this

If $y=a+bx+cx^2 \geq 0$ then $y|_{\min}\ge 0$. $$x_{min}=\frac{-b}{2c}$$ therefore $ y(x_{min})\ge0$. If we calculate $$y\left(\frac{-b}{2c}\right)$$ we will obtain ${b^2 - 4ca}\leq 0$.

but i dont know how to use it in this context.

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  • $\begingroup$ I put $\geq 0$ into your inequality based off of the context. $\endgroup$
    – rschwieb
    Feb 4, 2013 at 20:47

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I think you are overthinking this.

Plug the values for $a,b,c$ back into $b^2-4ac\leq 0$ and see what you have. (It is right in front of you!)

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  • $\begingroup$ okay if i plug the values in i obtain, $\endgroup$
    – MathGeek
    Feb 5, 2013 at 1:54
  • $\begingroup$ okay if i plug the values in i obtain, $2(\bf{u}\cdot\bf{v}) -4|\bf{v}|^2|\bf{v}|^2 = 2(\bf{u}\cdot\bf{v}) - 4\bf{v}\bf{v}\bf{u}\bf{u} = (\bf{u}\cdot\bf{v})[2-4(\bf{u}\cdot\bf{v})] \leq 0 $ but i dont know what that is suppose to show me? $\endgroup$
    – MathGeek
    Feb 5, 2013 at 2:06
  • $\begingroup$ $2(u\cdot v)$ is not $b^2$. $-4|v|^2|v|^2$ is not $-4ac$. Check again. $\endgroup$
    – rschwieb
    Feb 5, 2013 at 12:46

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