= Solution
For a <variation> $y+\varepsilon\eta$ preserving both endpoint values and slopes, $\eta=\eta'=0$ at the endpoints. The <first variation> is
$$
\delta J=\int_a^b(f_y\eta+f_{y'}\eta'+f_{y''}\eta'')\,dx.
$$
Integrate the last term twice and the middle term once. All boundary terms vanish, giving
$$
\delta J=\int_a^b\left(f_y-\frac{d}{dx}f_{y'}+\frac{d^2}{dx^2}f_{y''}\right)\eta\,dx.
$$
The <fundamental lemma of the calculus of variations> yields the <higher-order Euler-Lagrange equation>
$$
\boxed{f_y-\frac{d}{dx}f_{y'}+\frac{d^2}{dx^2}f_{y''}=0.}
$$
For the present functional this becomes $1+y-y^{(4)}=0$. Its general solution is $y=-1+A\cosh x+B\sinh x+C\cos x+D\sin x$. The left endpoint gives $C=-A$ and $D=-B$. The right endpoint then gives
$$
A(\cosh\pi+1)+B\sinh\pi=\cosh\pi+1,\qquad A\sinh\pi+B(\cosh\pi+1)=\sinh\pi.
$$
The coefficient <determinant> is $2(1+\cosh\pi)>0$, so $A=1$, $B=0$. \b[The stationary function is $y_*(x)=-1+\cosh x-\cos x$.]
For any admissible perturbation $\eta$, exact quadratic expansion, followed by the stationarity equation and integration by parts, gives
$$
J[y_*+\eta]-J[y_*]=\frac12\int_0^\pi\left(\eta^2-(\eta'')^2\right)dx.
$$
Apply the supplied <Poincaré inequality> first to $\eta$, then to $\eta'$, whose endpoint values also vanish:
$$
\int\eta^2\le\int(\eta')^2\le\int(\eta'')^2.
$$
Thus \b[$J[y_*+\eta]\le J[y_*]$ for every admissible perturbation], proving a global maximum through the <clamped second-variation inequality>.
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