Why does the minimum energy field configuration require the fields to be constant? I am having a hard time in understanding a well known statement always made in the context of field theory.
Background
Consider a classical real scalar field theory with Lagrangian density given by $$\mathcal{L}[\phi]=\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi+\frac{1}{2}\mu^2\phi^2-\frac{\lambda}{4!}\phi^4$$
The Hamiltonian density for this theory is given by a Legendre transform to be
$$\mathcal{H}[\phi,\pi]=\frac{1}{2}\pi^2+\frac{1}{2}(\vec{\nabla}\phi)^2-\frac{1}{2}\mu^2\phi^2+\frac{\lambda}{4!}\phi^4$$
and we are asked to find the field configuration of minimum energy.
Usually one says that since $\mathcal{H}[\phi,\pi]$ is composed of a quadratic (hence always positive) part which is $\frac{1}{2}\pi^2+\frac{1}{2}(\vec{\nabla}\phi)^2$ then $\mathcal{H}[\phi,\pi]$ is smaller when this part is zero.
Then one, in order to make this part vanish, assumes the field to be constant $\phi(x)=a \ \in \mathbb{R}$ and then minimizes the potential $$V[\phi]=-\frac{1}{2}\mu^2\phi^2+\frac{\lambda}{4!}\phi^4$$ now seen to be not anymore a functional of fields, but merely a $V: \mathbb{R} \to \mathbb{R}$ funcion which sends $a \mapsto V(a)$.
The question
My question is the following: How do we say that this is true?
Why can not there be a field configuration $\phi(x)=\chi(x)$ which is not  constant and therefore does not make the quadratic part vanish, but still gives a much smaller potential and therefore an overall a smaller energy?
Some more reasoning I did:
Thinking of the analogy in calculus, it seems to me that this statement is similar to the following:
"In order to find the minimum of the function $F(x)=f^2(x)+g(x)$ look first at which values of $\tilde{x}_i \in \mathbb{R}$ you have $f(\tilde{x}_i)=0$. Then among these choose the one for which you have the minimum of $g(x)$. Let say with no loss of generality that is is $x_1$. Then $x_1$ is the global minimum of the function $F(x)$."
This statement is of course false and it is not hard to find counterexamples, such as
$$F(x)=[(x-1)(x-2)]^2+e^x$$
In fact applying the statement above I have that $\tilde{x_1}=1$ and $\tilde{x}_2=2$ are the zeroes of the quadratic part. Therefore I look among them and find $F(1)=g(1)=e$ and $F(2)=g(2)=e^2$ as candidate minima. Therefore I would conclude saying $x=1$ is the minimum but this is clearly false since since taking $x=\frac{1}{2}$ gives $F(1/2)<F(1)$
What is wrong with my reasoning?
Note (for all of you that posses Peskin-Schroeder) the question arises from reading the statements in page 348.
 A: The key insight is that the field configuration that is constant and whose constant value minimizes the potential energy density simultaneously minimizes both the potential energy functional and the kinetic energy functional individually.  Let' see how this works:
Let $T$ and $V$ be the kinetic and potential functionals whose sum is the Hamiltonian;
\begin{align}
  T[\pi,\phi] &= \int d^3x \left[\frac{1}{2}\pi(x)^2 + \frac{1}{2}(\nabla\phi(x))^2\right] \\
   V[\phi] &= \int d^3x\left[-\frac{1}{2}\mu^2\phi(x)^2 + \frac{\lambda}{4!}\phi(x)^4\right] = \int d^3x\, \mathscr V(\phi(x))
\end{align}
Let $\phi_0$ be any real number that minimizes the potential density
\begin{align}
  \mathscr V(\phi_0)\leq \mathscr V(\phi), \qquad \text{for all $\phi\in\mathbb R$}
\end{align}
Let $\mathcal C$ denote the set of all admissible field configurations $\phi$, and let $\mathcal P$ be the set of all admissible momentum configurations $\pi$. Then I claim that if $\phi_*$ is the field configuration whose value is $\phi_0$ everywhere, then
\begin{align}
  H[\pi, \phi_*]\leq H[\pi, \phi], \qquad \text{for all $\phi\in\mathcal C$ and $\pi\in \mathcal P$}.
\end{align}
The proof goes as follows.  Notice that given any field configuration $\phi$, we have
\begin{align}
  \mathscr V(\phi_0) \leq \mathscr V(\phi(x)), \qquad \text{for all $x\in\mathbb R^3$}.
\end{align}
By integrating both sides of this inequality, we immediately find that $\phi_*$ is a minimum of the potential energy functional; $V[\phi_*] \leq V[\phi]$ for all $\phi \in \mathcal C$.  Next, notice that since $(\nabla\phi)^2$ is manifestly non-negative, given any $\pi\in P$ we have $T[\pi, \phi]\geq T[\pi, \bar \phi]$ for any $\bar\phi$ constant, and for all $\phi\in\mathcal C$.  In particular, $T[\pi, \phi_*]\leq T[\pi, \phi]$ for all $\pi\in \mathcal P$ and $\phi\in\mathcal C$. In other words, for any $\pi$, the configuration $\phi_*$ is also a minimum of the kinetic energy functional.  Putting these facts together, we find that
\begin{align}
  V[\phi_*] + T[\pi, \phi_*]\leq V[\phi] + T[\pi, \phi], \qquad \text{for all $\phi\in\mathcal C$ and $\pi\in \mathcal P$}.
\end{align}
as desired.
