# Vector Proof

• Aug 30th 2012, 11:48 PM
TaTzRE66
Vector Proof
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||

• Aug 31st 2012, 04:08 AM
Prove It
Re: Vector Proof
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...
• Aug 31st 2012, 11:41 AM
awkward
Re: Vector Proof
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,
$|\mathbf{b} - \mathbf{c}| \ge |\mathbf{b}| - |\mathbf{c}| > |\mathbf{a}|$