Vector Proof

**TaTzRE66** · August 30th 2012, 10:48 PM

Hi all.

I am stuck at this vector proof question and need help.

Suppose vectors a,b,c satisfy ||a|| > (1/2) + sqrt((1/4)+||c||)

and ||b|| = (||a||)^2

Prove that ||b-c|| > ||a||

Someone please help, thank you.

**Prove It** · August 31st 2012, 03:08 AM

You have $\displaystyle \begin{align*} |\mathbf{a}| > \frac{1}{2} + \sqrt{\frac{1}{4} + |\mathbf{c}|} \end{align*}$ and $\displaystyle \begin{align*} |\mathbf{b}| = |\mathbf{a}|^2 \end{align*}$ .

From the first inequality

$\displaystyle \begin{align*} |\mathbf{a}|^2 &> \left( \frac{1}{2} + \sqrt{\frac{1}{4} + |\mathbf{c}|} \right)^2 \\ |\mathbf{a}|^2 &> \frac{1}{4} + \sqrt{\frac{1}{4} + |\mathbf{c}|} + \frac{1}{4} + |\mathbf{c}| \\ |\mathbf{a}|^2 &> \frac{1}{2} + \sqrt{\frac{1}{4} + |\mathbf{c}|} + |\mathbf{c}| \\ |\mathbf{b}| &> \frac{1}{2} + \sqrt{\frac{1}{4} + |\mathbf{c}|} + |\mathbf{c}| \\ |\mathbf{b}| - |\mathbf{c} | &> \frac{1}{2} + \sqrt{\frac{1}{4} + |\mathbf{c}|} \\ |\mathbf{b} - \mathbf{c}| &> \frac{1}{2} + \sqrt{\frac{1}{4} + |\mathbf{c}|} \end{align*}$

I'm afraid I don't know how to continue...

**awkward** · August 31st 2012, 10:41 AM

We have
$|\mathbf{a}| - 1/2 > \sqrt{1/4 + |\mathbf{c}|}$
so (squaring),

$|\mathbf{a}|^2 - |\mathbf{a}| + 1/4 > 1/4 + |\mathbf{c}|$

hence
$|\mathbf{a}|^2 - |\mathbf{c}| > |\mathbf{a}|$

substituting $|\mathbf{b}| = |\mathbf{a}|^2$ ,

$|\mathbf{b}| - |\mathbf{c}| > |\mathbf{a}|$

But by the triangle inequality,
$|\mathbf{b} - \mathbf{c}| + |\mathbf{c}| \ge |\mathbf{b} - \mathbf{c} + \mathbf{c}| = |\mathbf{b}|$ ,
so
$|\mathbf{b} - \mathbf{c}| \ge |\mathbf{b}| - |\mathbf{c}|$

Consequently,
$|\mathbf{b} - \mathbf{c}| \ge |\mathbf{b}| - |\mathbf{c}| > |\mathbf{a}|$

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