= Solution
Set $l=m=g=1$. At $(\theta_1,\theta_2)=(0,0)$ the <spring> has $d=r$, so its <potential energy> has zero <gradient>: $\partial_jV_s=k(d-r)\partial_jd=0$. The gravitational <potential energy> has $\partial_jV_g=\sin\theta_j=0$ there. Zero <velocities> therefore satisfy the <Euler-Lagrange equations>, proving this is an <equilibrium>.
For fixed $r>0$, let $\varepsilon=\max(|\theta_1|,|\theta_2|)$ and $\delta=\theta_2-\theta_1$. The <Taylor series> give
$$
\sin\theta_2-\sin\theta_1=\delta+O(\varepsilon^3),\qquad
\cos\theta_2-\cos\theta_1=O(\varepsilon^2).
$$
Thus, near the downward <equilibrium>, the horizontal separation is positive and
$$
d=\sqrt{\bigl(r+\delta+O(\varepsilon^3)\bigr)^2+O(\varepsilon^4)}
=r+\delta+O(\varepsilon^3).
$$
In particular, $V_s=k\delta^2/2+O(\varepsilon^4)$ and $V_g=-2+(\theta_1^2+\theta_2^2)/2+O(\varepsilon^4)$. Dropping the additive constant and terms beyond quadratic order gives the <harmonic approximation>
$$
\boxed{L_{\mathrm{quadratic}}=\frac12(\dot\theta_1^2+\dot\theta_2^2)
-\frac12\left(\theta_1^2+\theta_2^2+k(\theta_1-\theta_2)^2\right).}
$$
For physical <spring constant> $k\geq0$, the quadratic <potential energy> is a <positive-definite quadratic form>. The downward position is therefore a <stable equilibrium in Lagrangian mechanics>, and its nearby motion admits <small oscillations>.
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