# Why a system should be at its lowest energy state for its stability?

To answer your question, you should first understand when is a system most stable.

Firstly it shouldn't have a tendency to move or change state, thus it should be under equilibrium conditions, i.e. the net Force should be zero.

We know that $$F = - \frac{dU}{dx}$$ Putting $F=0$, we get $$\frac{dU}{dx}=0 \tag{1}$$

Secondly, it should be able to maintain that equilibrium condition by itself. This can be tested by displacing the system by a small distance $\delta x$. If the force on the system then becomes opposite to direction of $\delta x$, we can say that the system has a tendency to restore back to its original equilibrium position.
An example of this would be a ball kept at the bottom of a spherical valley. Displace the ball a little towards the right, and the net force on it acts towards left, bringing it back to its original position. You will realise that I just described a stable equilibrium condition. What this proves is that it is the stable equilibrium condition in which the system is most stable.

From the above description we have that the small displacement $\delta x$ and net extra force $\delta F$ should be in opposite directions

$$\delta F = \frac{dF}{dx} \delta x + \mathcal{O}(\delta x^2) \approx \frac{dF}{dx} \delta x$$

which gives as stability condition

$$\frac{dF}{dx} < 0$$

which implies

$$-\frac{d^2U}{dx^2}<0$$ $$\frac{d^2U}{dx^2}>0\tag{2}$$

From $(1)$ and $(2)$ it is evident that the graph of $U$ should have a minima at stable equilibrium condition, i.e. The Potential Energy should be minimum when a system attains maximum stability.