= Solution
Put $\epsilon_k=\hbar^2k^2/(2m)>0$ for $\mathbf k\ne0$ and $g=u\rho$, using $\rho=N/L^3$. Since the zero mode has zero <kinetic energy>, the exact <kinetic energy> in this trial state is
$$
\boxed{\langle H_{\rm KE}\rangle
=\sum_{\mathbf k\ne0}\epsilon_k\sinh^2\theta_k.}
$$
Together with part (d), the <fixed-mean-number Bogoliubov variational energy> is
$$
\mathcal E[\theta]\simeq\frac{gN}{2}
+\sum_{\mathbf k\ne0}f_k(\theta_k),\qquad
f_k(\theta)=(\epsilon_k+g)\sinh^2\theta-g\sinh\theta\cosh\theta.
$$
For repulsive <contact interaction>, $u>0$, each $\epsilon_k+g>g>0$. Using the double-angle <hyperbolic-function identities>,
$$
f_k(\theta)=\frac{\epsilon_k+g}{2}(\cosh2\theta-1)
-\frac g2\sinh2\theta,
$$
so
$$
f_k'(\theta)=(\epsilon_k+g)\sinh2\theta-g\cosh2\theta.
$$
The <stationary point> therefore satisfies
$$
\boxed{\tanh2\theta_k=\frac{g}{\epsilon_k+g}
=\frac{u\rho}{u\rho+\hbar^2k^2/(2m)}.}
$$
This is a minimum, not just a vanishing <derivative>. Define the positive <Bogoliubov quasiparticle dispersion>
$$
E_k=\sqrt{(\epsilon_k+g)^2-g^2}
=\sqrt{\epsilon_k(\epsilon_k+2g)}.
$$
At the displayed solution $\theta_k^*$, $\cosh2\theta_k^*=(\epsilon_k+g)/E_k$ and $\sinh2\theta_k^*=g/E_k$. Hence
$$
f_k(\theta)=\frac{E_k}{2}\cosh[2(\theta-\theta_k^*)]
-\frac{\epsilon_k+g}{2},\qquad
f_k''(\theta_k^*)=2E_k>0.
$$
The <hyperbolic cosine> has its unique minimum at zero, proving global minimization of each retained term. Treating equal-magnitude <wavevectors> together simply multiplies the same function by their multiplicity. At a fixed <ultraviolet cutoff>, the resulting <energy> change from $\theta=0$ is
$$
\boxed{\mathcal E_{\min}-\mathcal E[0]
=\frac12\sum_{\mathbf k\ne0}(E_k-\epsilon_k-g)<0\quad(g>0).}
$$
Indeed $E_k<\epsilon_k+g$. More directly, for small positive $\theta$,
$$
f_k(\theta)=-g\theta+(\epsilon_k+g)\theta^2+O(\theta^3).
$$
The pair <vacuum expectation value> lowers the <interaction energy> at first order in $\theta$, whereas the nonzero <occupation number> and its <kinetic energy> cost start at second order. The <two-mode squeezed vacuum> admixes opposite-<momentum> pairs into the condensate, introducing spatial correlations that reduce the chance of two <bosons> coinciding. These correlations lower repulsive <contact interaction> enough to outweigh the added <kinetic energy>. A small <quantum depletion of a Bose condensate> can therefore lower the <energy> below the uncorrelated condensate value.
Two qualifications delimit this result. First, the paper does not state the sign of $u$. More generally a finite stable minimum exists only when $\epsilon_k+g>|g|$ for every retained nonzero mode; repulsion ensures this. For example, with $g=-\epsilon_k$ in some mode, $f_k(\theta)=\epsilon_k\sinh(2\theta)/2$ tends to $-\infty$ as $\theta\to-\infty$, so an unconditional minimum claim for arbitrary $u$ is false. Also the formula minimizes the truncated <Bogoliubov theory> <energy>, not the exact quartic <Hamiltonian> including its discarded terms, and it must remain in the regime of small <quantum depletion of a Bose condensate>.
Second, a three-dimensional unregulated <contact interaction> cannot give a finite absolute beyond-leading-order <ground-state energy> by this bare sum. For large $k$, $E_k-\epsilon_k-g=-g^2/(2\epsilon_k)+O(\epsilon_k^{-2})$, whose three-dimensional sum has an <ultraviolet divergence>. A finite <ultraviolet cutoff>, or <renormalization of a three-dimensional dilute-gas contact coupling> matched to a physical <scattering length>, is needed. This does not alter the modewise minimization or the finite-cutoff lowering mechanism just derived.
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