What are the Values of x, y, z for Mechanical Energy to be Positive?

[SpY]]

Homework Statement

A particle of mass m experiences a force
\overrightarrow{F} = k (-x\hat{i} -y\hat{j} -z\hat{k})

Find values of x, y, z for which the mechanical energy E > 0.

Homework Equations

Conservation of mechanical energy
\frac{1}{2}mv^2 + V = E = constant
where \frac{1}{2}mv^2 is the kinetic energy and V the potential energy of the system. (This is only valid in an isolated system, ie. no external forces)

The Attempt at a Solution

First checked that the force was conservative (curl is zero), then used a line integral to find a potential function.
I get V = \frac{1}{2}kz^2.
Then using the conservation law I get \frac{1}{2}mv^2 + \frac{1}{2}kz^2 = E > 0

From there I'm not sure what to do. I tried decomposing the velocity,
\overrightarrow{v} = \dot{x}\hat{i} + \dot{y}\hat{j}+ \dot{z}\hat{k}
Then get 3 differential 'inequalities'. I'm not sure, but is it valid to say \dot{x}^2 = 2\ddot{x}?
Just confirming this approach before going on

 

[olivermsun]

Double-check your potential function. Does \mathbf{F} = -\nabla \Phi for the function you got?

 

[SpY]]

! Yes it was wrong. Here's my second try:
V = -\int_{0,0,0}^{x,0,0}-kx dx + -\int_{x,0,0}^{x,y,0}-ky dy + -\int_{x,y,0}^{x,y,z}-kz dz
V = \frac{k}{2}(x^2+y^2+z^2) +C
verifying
-\nabla V = -\frac{\partial }{\partial x} \frac{kx^2}{2} -\frac{\partial }{\partial y} \frac{ky^2}{2} -\frac{\partial }{\partial z} \frac{kz^2}{2}
= -k(x\hat{i}+y\hat{j}+z\hat{k}) = \overrightarrow{F}

Then the conservation law is
\frac{1}{2}m\left |\overrightarrow{v} \right |^2 + \frac{k}{2}(x^2+y^2+z^2) > 0

|\overrightarrow{v}| ^2 > -\frac{k}{m}(x^2+y^2+z^2)

Then square root and let \overrightarrow{v} = \dot{r} \;\;\; \text{and} \;\; r = \sqrt{x^2+y^2+z^2}

\dot{r} + \sqrt{-\frac{k}{m}}r > 0

Simplest solution I get is \displaystyle{r > Ce^{-\gamma t} \; \; \text{where} \; \; \gamma = \sqrt{-\frac{k}{m}}}

If k>0 then the real part, by Euler's formula is r > Acos(\sqrt{\frac{k}{m}} t)