= Solution
At zero, the sixth-power term has zero second <derivative>, so
$$
\delta^2I_\lambda(0)(\varphi,\varphi)=2Q_\lambda(\varphi),\qquad Q_\lambda(u)=\int_0^\pi(u'^2-\lambda^2u^2)dx.
$$
For $0<\lambda<1$, the Jacobi solution $z(x)=\cos(\lambda(x-\pi/2))$ is strictly positive on $[0,\pi]$. The preceding identity gives $Q_\lambda(u)=\int z^2[(u/z)']^2>0$ for nonzero zero-trace $u$.
To control the nonlinear remainder, a uniform lower bound follows directly. Let $m=\min z>0$, $K=\max|z'|$, and $v=u/z$. Since $\max z=1$ and $\int v^2\leq(\pi^2/2)\int v'^2$ by the elementary endpoint estimate,
$$
\int u'^2\leq2K^2\int v^2+2\int v'^2\leq(K^2\pi^2+2)\int v'^2.
$$
Thus $Q_\lambda(u)\geq c_\lambda\int u'^2$ with $c_\lambda=m^2/(K^2\pi^2+2)>0$. Also $\int u^6\leq\|u\|_\infty^4(\pi^2/2)\int u'^2$, so
$$
I_\lambda(u)\geq\left(c_\lambda-\frac{\pi^2}{4018}\|u\|_\infty^4\right)\int u'^2.
$$
This is positive for every nonzero admissible $u$ sufficiently close to zero in $W^{1,\infty}$. Therefore \b[zero has positive <second variation> and is a strict weak local minimizer]. No sharp <Poincaré inequality> has been assumed in this proof.
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