Question and Conjecture
In a Lorentzian metric (Only one time dimension) of signature: $(+,-,-,-,\dots)$
Lets say I have a line element $ds^2 = \sum_{\mu , \nu = 0}^n g^{\mu \nu} d \tilde x_\mu d \tilde x_\nu $ and
$1.$ It is of the form $ds^2 = \sum_{\mu=0}^n g^{\mu \mu} d \tilde x_\mu d \tilde x_\mu$
$2.$ There exists a continuous path parametrized by $\lambda$ whose path-length extends to infinity
We define $s = \int_{\lambda_0}^{\lambda_1} \sqrt{\sum_{\mu=0}^n \frac{d \tilde x_\mu}{d \lambda} \frac{d \tilde x^\mu}{d\lambda}} d \lambda $ with $s(0)= 0 = \lambda_0$
$3.$ It must be of the Lorentzian only $g_{00}$ must be positive and the path must be timelike. Then:
We define:
$$ g^{00} ds'^2 = ds^2 $$
and define coordinates: $d x_i = \sqrt{|g^{ii}|} d \tilde x_i$. Then:
$$(\int_{s(\lambda_0)=0}^\infty e^{-\alpha s'^2} ds' )^2 \geq (\int_{\lambda_0}^\infty e^{-\alpha s(x (\lambda),y(\lambda),z(\lambda))^2} \frac{dx_0}{d \lambda} d \lambda )^2 - \sum_{\mu=1}^n (\int_{\lambda_0}^\infty e^{-\alpha s(x(y),y,z(y))^2} \frac{d x_\mu}{d \lambda} d \lambda )^2 $$
Are there any counter examples?
Heuristic Rough Derivation and Example
The following proof is difficult to follow but is meant to give only a rough sketch.
We define $s$ as done already previously now let's say I have the line element:
$$ ds^2 = f(r,t)^2 dt^2 - r^2 d\theta^2 - dr^2 $$
We compare this to:
$$ f(r,t) ds' = ds $$
Hence,
$$ ds'^2 = dt^2 - \frac{dr^2}{f(r,t)^2} - \frac{r^2}{f(r,t)^2} d \theta^2 $$
Now we multiply a factor of $\beta_s e^{-\alpha s'(x,y,z)}$ both sides. Note: $\beta_s$ should not be thought as a function
$$ \beta_s e^{\alpha s'} ds'^2 = e^{\alpha s'} \beta_s dt^2 - \beta_s e^{\alpha s'} \frac{dr^2}{f(r,t)^2} - \beta_s e^{\alpha s'} \frac{r^2}{f(r,t)^2} d \theta^2 $$
Let:
$$ \beta_s e^{\alpha s'} ds'^2 = e^{\alpha s'} \beta_s dt^2 - \beta_s e^{\alpha s'} \frac{dr^2}{f(r,t)^2} - \beta_s e^{\alpha s'} \frac{r^2}{f(r,t)^2} d \theta^2 $$
We can also say:
$$\beta_s e^{\alpha s'} ds'^2 \geq e^{\alpha s'} \beta_s dt^2- \beta_s e^{\alpha s'} (d \int \frac{ dr}{f(r,t)} )^2 - \beta_s e^{\alpha s'} (d \int \frac{r}{f(r,t)} d \theta)^2 $$
We define $y = \int \frac{ dr}{f(r,t)} $ and $z = \int \frac{r}{f(r,t)} d \theta$
Now we assume the path takes is continuous. We then do a transformation based on this and cancel the common coefficient:
$$(\int_{s(\lambda_0)=0}^\infty e^{-\alpha s'^2} ds' )^2 \geq (\int_{\lambda_0}^\infty e^{-\alpha s(x (\lambda),y(\lambda),z(\lambda))^2} \frac{dt}{d \lambda} d \lambda )^2 $$ $$- (\int_{y(\lambda_0)}^\infty e^{-\alpha s(x(y),y,z(y))^2} \frac{dy}{d \lambda} d \lambda )^2 - (\int_{\lambda_0}^\infty e^{-\alpha s(x(z),y(z),z))^2} \frac{dz}{d \lambda} d \lambda )^2 $$
