Killing vectors I have an assignment:

For a metric $g_{\mu\nu}$ with everywhere timelike Killing vector  $K^\mu$, a free particle with $p^\mu$ and mass $m$ show that its conserved energy $E=-p_\mu K^\mu$ is bound from below by $m\sqrt{-K_\mu K^\mu}$.

I can easily show it for a unit vector (1, 0, 0, 0) (we can go move into its system), show that it is indeed a Killing vector by satisfying $\nabla_{(\mu} K_{\nu)}=0$ and write scalar product which is $-1$, then its energy is bound from below by $m$, but I can't show it generally.
How can I proceed?
 A: Very nice, I did not know!
The fact that $K$ is a Killing vector does not matter here. It only matters that both $p$ and $K$ are timelike and future directed. The thesis arises from the statement
Proposition.
If $A$ and $B$ are future directed timelike vectors, then the inverse Cauchy-Schwartz inequality holds: $$-A_\mu B^\mu \geq \sqrt{-A_\mu A^\mu} \sqrt{-B_\mu B^\mu}\:.$$
Proof.
$$-A_\mu B^\mu = A^0B^0 - \vec{A}\cdot \vec{B} \geq A^0B^0 - |\vec{A}\cdot \vec{B}| \geq  A^0B^0 - ||\vec{A}||||\vec{B}|| \geq \sqrt{((A^0)^2- ||\vec{A}||^2)((B^0)^2- ||\vec{B}||^2)}= \sqrt{-A_\mu A^\mu} \sqrt{-B_\mu B^\mu}\:.$$
The last inequality is immediate using the fact that $A^0, B^0 >0$ and proving it into the equivalent version
$$(A^0B^0 - ||\vec{A}||||\vec{B}||)^2 \geq ((A^0)^2- ||\vec{A}||^2)(B^0)^2- ||\vec{B}||^2)$$
whose proof is immediate because it boils down to
$$(A^0||\vec{B}||-B^0||\vec{A}||)^2\geq 0\:.$$ $ \Box$
Finally we observe that $-p_\mu p^\mu = m^2$ and it concludes the proof, using $A=p$ and $B=K$.
PS. There is an even  shorter proof. Arrange a pseudo-orthonormal basis with $e_0 \parallel A$. In this frame $-A_\mu B^\mu = A^0B^0 = \sqrt{-A_\mu A^\mu}\sqrt{(B^0)^2} \geq \sqrt{-A_\mu A^\mu}\sqrt{(B^0)^2- ||\vec{A}||^2} = \sqrt{-A_\mu A^\mu}\sqrt{-B_\mu B^\mu}$.
The result is valid in general because the left-most and the right-most terms are invariant.
