= Solution
With unit constant bending coefficients and unit length, the quadratic <elastic energy> is
$$
E=\frac12\int_0^1\left(y''^2+z''^2-F(y'^2+z'^2)\right)\,dx.
$$
To find the first loss of positivity, put $v=y'$. The clamped conditions give $v(0)=v(1)=0$ and $\int_0^1v\,dx=0$. Extend $v$ periodically with period one. The sharp <periodic Wirtinger inequality> gives $\int_0^1v'^2\,dx\ge4\pi^2\int_0^1v^2\,dx$. For completeness, expand the mean-zero periodic $v$ in its <Fourier series> $\sum_{n\ge1}(a_n\cos2\pi nx+b_n\sin2\pi nx)$. <Parseval identity> gives
$$
\int_0^1v^2\,dx=\frac12\sum_{n\ge1}(a_n^2+b_n^2),\qquad
\int_0^1v'^2\,dx=\frac12\sum_{n\ge1}(2\pi n)^2(a_n^2+b_n^2),
$$
which proves the inequality and its equality condition. The same holds for $z'$. Consequently
$$
E\ge\frac{4\pi^2-F}{2}\int_0^1(y'^2+z'^2)\,dx.
$$
For $F<4\pi^2$, the straight filament is the unique minimum. Equality in the <periodic Wirtinger inequality> has $v=a\cos2\pi x+b\sin2\pi x$; the endpoint value $v(0)=0$ eliminates $a$. Integration and the position boundary conditions therefore give \b[the first buckling load and modes]:
$$
\boxed{F_c=4\pi^2,\qquad y=\alpha(1-\cos2\pi x),\quad z=\beta(1-\cos2\pi x).}
$$
At this load every such pair has zero quadratic <elastic energy> and solves the <Euler-Lagrange equations>. For $F>F_c$ these modes give negative energy, signaling instability of the straight configuration. The quadratic small-slope model does not select a finite post-buckling amplitude; nonlinear terms would be needed for that. This is <Euler buckling of an elastic filament>.
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