In deriving the Euler equation for etremizing a functional
\begin{equation*} J[y] = \int_a^b F(x,y,y')\,dx, \end{equation*}
we look at:
\begin{equation*} J[y+h]-J[y] = \int_a^b(F_yh+F_{y'}h')\,dx + \int_a^b(\text{higher order terms in $h$ and $h'$})\,dx \end{equation*}
For $J$ to have, say a minimum at $y$, the integral of the linear terms must be zero. Otherwise, the $\Delta J[h]$ can change sign if $h$ is replaced with $-h$.
This argument is valid only for $||h||_1<\epsilon$, $\epsilon$ being a very small positive number. I was wondering what happens when $||h||_1$ becomes larger. Does the non-linear part increase? Does the linear part increase, and if so, was the condition that $\int_a^b(F_yh+F_{y'}h')\,dx=0$ only a limit as $||h||\rightarrow 0$? Also, if it was a limit, does the argument of alternating sign in the paragraph above still hold true?
Maybe my answer will look like cheating, but it is not. Consider $J \colon (a,b) \to \mathbb{R}$, and assume $y \in (a,b)$ is a minimum point. Then $$ J[y+h]-J[y] = J'[y]h + \ldots \tag{1} $$ where the dots denote higher order terms in $h$. Now the linear term must vanish, as you noticed. What happens if $h$ becomes larger and larger? Anything may happen, we don't know.
The same happens for functionals: we don't know, maybe that functional increases, maybe the functional oscillates... Moreover, remark that the linear term is actually... linear in $h$: the "size" of $h$ does not really matter, since any $h \neq 0$ can be normalized.
By the way, you are just saying that the (Fréchet) derivative must vanish at a minimum point, nothing more.