The question and my attempt can be found here: http://i.imgur.com/CiMNr2m.jpg?1

I don't quite understand what i'm suppose to do. It says to prove the inequality with whatever so I tried to substitute and then factor and I thought it would just work out, but it didn't.


  • $\begingroup$ What you have written in the picture says you're supposed to prove the equality $\endgroup$ – Tyler Oct 2 '13 at 0:55
  • $\begingroup$ So that means, that I use = signs instead of <= signs? $\endgroup$ – Kat Oct 2 '13 at 1:08

Your work looks good. I think you are making things a bit too hard for yourself. I'll compute one side for you; then, hopefully, you can verify that the two sides are indeed equal. If $x_{1} = \lambda y_{1}$ and $x_{2} = \lambda y_{2}$, then we have:

$$ \sqrt{x_{1}^{2}+x_{2}^{2}}\sqrt{y_{1}^{2}+y_{2}^{2}} =\sqrt{(\lambda y_{1})^{2}+(\lambda y_{2})^{2}}\sqrt{y_{1}^{2}+y_{2}^{2}}$$ Then, factoring out a $\lambda^{2}$, we find:

$$ \sqrt{x_{1}^{2}+x_{2}^{2}}\sqrt{y_{1}^{2}+y_{2}^{2}} =\lambda\sqrt{y_{1}^{2}+ y_{2}^{2}}\sqrt{y_{1}^{2}+y_{2}^{2}} = \lambda ((y_{1}^{2}+ y_{2}^{2})^{1/2})^{2}$$

But we can see that this is just

$$\lambda(y_{1}^{2} + y_{2}^{2})$$

  • $\begingroup$ when factoring out the lambda, how come its ok to factor even though there is a sqrt(y_1^2+ y_2^2)? $\endgroup$ – Kat Oct 2 '13 at 1:17
  • $\begingroup$ If I understand your question correctly, I am using the general result that for positive $a, b$, we have $\sqrt{a^{2}b} = \sqrt{a^{2}}\sqrt{b} = a\sqrt{b}$. $\endgroup$ – Alex Wertheim Oct 2 '13 at 1:24

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