= Solution
Here $P=2$, $Q=-2\lambda^2$ and the <Jacobi equation> is $w''+\lambda^2w=0$. With $w(0)=0$ and $w'(0)=1$, its solution is $w(x)=\sin(|\lambda|x)/|\lambda|$ for $\lambda\ne0$, and $w(x)=x$ for $\lambda=0$. Its first positive zero is $\pi/|\lambda|$ when $\lambda\ne0$.
If $\lambda^2<\pi^2/b^2$, no <conjugate point> occurs in $(0,b]$. The <Jacobi positivity criterion> gives
$$
\boxed{\delta^2I_\lambda(0)(\varphi,\varphi)=2\int_0^b(\varphi'^2-\lambda^2\varphi^2)dx>0\quad(\varphi\ne0).}
$$
Since $I_\lambda$ itself is this <quadratic form> divided by two, $I_\lambda(u)>0=I_\lambda(0)$ for every nonzero admissible $u$. Thus \b[zero is the unique <global minimizer>]. This argument establishes the result before using the sharp <Poincaré inequality>.
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