= Solution
{c}
Use the same normalization for the <elastic energy> and electronic <energy density>: if $\kappa\delta^2$ is an <energy> per <unit cell>, divide it by $a$ first; if it is already per length, use it directly. Write its coefficient per length as $K>0$ and put $A=\pi^2V_0^2/(4aE_{\rm ZB})>0$. The small-distortion change in total <energy density> is
$$
\mathcal U_{\rm tot}(\delta)-\mathcal U_{\rm tot}(0)
=\delta^2\left[K-A+2A\ln\frac{C|\delta|}{4}\right]+O(\delta^4).
$$
No minimization calculation is needed. For arbitrarily small nonzero $|\delta|$, the <logarithm> is negative with arbitrarily large magnitude. Hence the negative electronic <Peierls energy gain>, proportional to $\delta^2\ln(1/|\delta|)$, eventually exceeds any fixed positive quadratic <elastic energy>. There are distorted configurations with lower <energy> than the undistorted configuration for every finite $K$ and nonzero coupling. The ideal zero-<temperature> infinite-chain model therefore has
$$
\boxed{|\delta_{\rm eq}|>0.}
$$
The source's $\delta_{\rm eq}>0$ is correct when $\delta$ denotes a nonnegative distortion amplitude. If $\delta$ is signed, the <energy density> is even and the two signs describe equally stable patterns related by a <lattice translation>. For example, $V(x+a/2;\delta)=V(x;-\delta)$, so a negative equilibrium distortion has the same <energy> as the positive one. The zero-coupling case $V_0=0$ instead has its minimum at $\delta=0$. The argument assumes a coupled chain, complete lower-<energy band> filling and zero <temperature>; it does not by itself establish finite-<temperature> <long-range order> in an isolated one-dimensional system.
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