Time-energy uncertainty relation shows the universe is everlasting

Heisenberg’s time-energy uncertainty relation is given by the below equation:
Δt ΔE ≥ ℏ/2

While commenting in a blog by German physicist Sabine Hossenfelder, another physicist has written the following in one of his comments in the comment section:
“Small ΔE means long lifetime.”1

My starting point will be this: Small ΔE means long lifetime.

If small ΔE means long lifetime, then, what will have to be the lifetime if ΔE is to be zero?

We have already seen that Δt ΔE ≥ ℏ/2.

From the above equation, we can write:
ΔE ≥ ℏ/2 multiplied by 1/Δt

For ΔE to be zero, the other side of the equation will also have to be zero. But ℏ/2 cannot be zero, because as per the time-energy uncertainty relation, the product of ΔE and Δt must always have a value greater than or equal to ℏ/2. So, for ΔE to be zero, 1/Δt must be zero. 1/Δt will be zero if, and only if, value of Δt is infinite. This is because inverse of infinity is zero. This shows that only an entity having an infinitely long lifetime can have zero energy.

But, we also know that the total energy of the universe is zero. That means the universe has an infinitely long lifetime; that means the universe is everlasting.

The universe as a whole is everlasting. But whatever is there within the universe has a limited lifetime only; none of them is everlasting. Having a limited lifetime only, none of them can have zero energy. Having a limited lifetime only, they will all have a non-zero energy value. That is the reason as to why even a perfect vacuum within the universe is also having energy, which is called vacuum energy.

Does the above make any sense?

1. Sabine Hossenfelder’s Blog How do black holes destroy information and why is that a problem? Comment by PhysicistDave 6:41 AM, August 28, 2019