Let $f$ be a positive, nondecreasing, differentiable real function
We want to prove the following inequality: $$\int_{0}^{\infty}e^{-t-f(t)}\sqrt{1+\left ( f'(t) \right )^2}dt\geq \sqrt{\alpha ^2+(\alpha -e^{-f(0)})^2}$$ where $$\alpha =\int_{0}^{\infty }e^{-t-f(t)}dt$$
We observe that, since $f$ is nondecreasing and positive, the following limit must exist and it must be equal to either $$\lim_{t\rightarrow \infty }f(t) = l>0$$ or $$\lim_{t\rightarrow \infty }f(t) =+\infty$$
In both cases it follows that $$\lim_{t\rightarrow \infty}e^{-t-f(t)}=0$$ so the inequality we want to prove looks something like this $$\int_{0}^{\infty}e^{-t-f(t)}\sqrt{1+\left ( f'(t) \right )^2}dt\geq \sqrt{(\alpha -e^{-\infty-f(\infty)})^2+(\alpha -e^{-0-f(0)})^2}$$
So we have some sort of symmetry. The RHS looks like a distance, but I cannot figure out the role that $\alpha$ plays in this "distance". For the LHS we observe that it looks like the line integral of the function $$z(x,y)=e^{-x-y}$$ for $x=0$ to $x\rightarrow\infty$
I have not been able to proceed further than these observations. One last thing I thought of was writing the integral as $$\int_{0}^{\infty}e^{-t-f(t)}\sqrt{1+\left ( f'(t) \right )^2}dt=\lim_{M\rightarrow\infty}\int_{0}^{M}e^{-t-f(t)}\sqrt{1+\left ( f'(t) \right )^2}dt$$ and then use the Chebyshev or Hölder inequalities to break the integral into a product of integrals, since we already have an expression for the first part, but didn't get to a solution