What is the Triangle Inequality and How Does Vector Normalization Work?

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http://img405.imageshack.us/img405/8026/prover.jpg

I would say it's pretty weak, how could I improve upon it?

It's supposed to be the triangle inequality.
 
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I don't mean to be rude, but frankly: it's not a proof at all. It's just two columns with a list of things that seem to be unrelated, and something that looks like what you wanted to prove at the bottom.

How about you use the definition of norm, as
|\vec{v}|^2 = \vec v \cdot \vec v
and work out
|\vec{u} + \vec{v}|^2
 
I'm just going off of the material presented in the chapter of a calc/linear algebra book, it is the first chapter and hasn't touched on anything further.

\overline{A} \overline{B} \ + \ \overline{B} \overline{C} \ = \ \overline{AC}

if I'm not mistaken, then I went to normalize the vectors but I think I got it wrong.

I'll just keep reading...
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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