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We prove the stated obstruction. All lengths are Euclidean arc lengths; crossing points are counted once, which has no effect on length integrals.

The family and its limiting line

For $y>0$ put

$$F_y(z)=z^4+(2y^2-\tfrac32)z^2-(1+4y^2)z.$$

Its derivative is $4(z-1)((z+\frac12)^2+y^2)$, so its three simple critical points are $1$ and $-\frac12\pm iy$. Their values are

$$v_y=-\tfrac32-2y^2,\qquad u_y\pm4iy^3,\qquad u_y=\tfrac3{16}+\tfrac52y^2-y^4.$$

Thus $d_y=v_y-u_y=y^4-\frac92y^2-\frac{27}{16}$. Its positive zero is $Y=\sqrt{\frac94+\frac32\sqrt3}$, and $d_y>0$ for $y>Y$. The circle $\Gamma_y$ through these values has real center and radius

$$C_y=v_y-R_y,\qquad R_y=\frac{d_y^2+16y^6}{2d_y}.$$

Indeed $|v_y-C_y|=R_y$ and $(u_y-C_y)^2+16y^6=(R_y-d_y)^2+16y^6=R_y^2$. Consequently $p_y(z)=R_y^{-1}F_y(R_y^{1/4}z)-C_y/R_y$ is centered and monic and has three distinct critical values on the unit circle. Its critical points are distinct too.

As $y\downarrow Y$, the radius tends to infinity and $-C_y/R_y=1-v_y/R_y\to1$. The other two nonleading coefficients tend to zero. Hence $p_y\to z^4+1$. On every bounded set $\Gamma_y$ converges smoothly to the line

$$L=-c+i\mathbb R,\qquad c=6+3\sqrt3,$$

and $F_Y(z)=z^4+(3+3\sqrt3)z^2-(10+6\sqrt3)z$ has all three critical values on $L$.

The limiting integral is positive

For any centered monic quartic $F$, set

$$G_F(\sigma)=\sum_{F(z)=\sigma}\frac1{|F'(z)|}$$

away from its critical values. Write $F=F_Y$ and define

$$J=\int_{-\infty}^{\infty} \left(G_F(-c+iv)-|v|^{-3/4}\right)\,dv, \qquad I=\int_{-\infty}^{\infty} \left(G_F(-c+iv)-(c^2+v^2)^{-3/8}\right)\,dv.$$

Both differences are absolutely integrable. Near a critical value the inverse branches have integrable square-root singularities, since the critical points are simple. At infinity the estimate proved below gives $G_F(\sigma)=|\sigma|^{-3/4}+O(|\sigma|^{-7/4})$, and translating the reference by the fixed number $c$ preserves integrability.

Let $H=\{z:\operatorname{Re}F(z)=-c\}$. Centering and monicity give eight ends of $H$, with asymptotic directions

$$e_j=\exp\bigl(i(\pi/8+j\pi/4)\bigr),\qquad 0\le j<8.$$

For sufficiently large $r$ each end meets $|z|=r$ once, at $re_j+O(r^{-1})$. This follows by applying the implicit function theorem to $\operatorname{Re}(r^{-4}(F(re^{i\theta})+c))=0$ near the eight simple zeros of $\cos4\theta$; differentiating also makes the end radial coordinate eventually monotone. The same expansion shows

$$J=\lim_{r\to\infty}\bigl(|H\cap\{|z|\le r\}|-8r\bigr). \tag{1}$$

For completeness, cutting the integral defining $J$ at $|v|\le r^4$ gives the length of the part of $H$ with $|F(z)+c|\le r^4$, minus $8r$. Along each end its cutting point differs from that on $|z|=r$ by $O(r^{-1})$, because $F$ has no cubic term. Thus the two length cuts differ by $O(r^{-1})$.

The only vertices of $H$ are its three critical points $w_0=1$, $w_+=-\frac12+iY$, $w_-=-\frac12-iY$, each of valence four. There is no cycle: a cycle would contain a Jordan curve on which the harmonic function $\operatorname{Re}F+c$ vanishes, forcing that function to vanish on the enclosed domain by the maximum principle. Cut the eight ends at a sufficiently large circle and regard the cut points as vertices of valence one. The resulting finite forest has eleven vertices and ten edges, by the degree sum $3\cdot4+8=20$. It is therefore a tree. In particular it has two finite edges connecting the three critical points. Their total length is at least

$$2\sqrt{\tfrac94+Y^2}>5, \tag{2}$$

since the two distances from $w_0$ to $w_\pm$ are the two shortest pairwise distances and $Y>2$.

For a unit vector $e$ let $h(e)=\max_{w\in\{w_0,w_+,w_-\}}\operatorname{Re}(w\bar e)$. An end beginning at a critical point has length, up to its intersection with $|z|=r$, at least $r-h(e_j)+o(1)$. In the upper half-plane the maximizing point is $w_0$ in direction $\pi/8$, and $w_+$ in directions $3\pi/8,5\pi/8,7\pi/8$. To check the first two assertions, use $Y<9/4<\frac32\cot(\pi/8)$ and $Y>2>\frac32\cot(3\pi/8)$; the last two are immediate. Conjugation handles the lower half-plane. Writing $q=\cos(\pi/8)$ and $s=\sin(\pi/8)$, we obtain

$$\sum_{j=0}^7h(e_j)=3q+2Y(2q+s) <3+\frac92\left(2+\frac12\right)=\frac{57}{4}. \tag{3}$$

Here $2<Y<9/4$, $q<1$ and $s<1/2$. The bound on $Y$ follows directly from its displayed square: $Y^2>4$ and $Y^2<81/16$. Combining (1)--(3) gives $J>-37/4$.

The difference between the two reference integrals is exactly

$$I-J=c^{1/4}K,\qquad K=\int_{-\infty}^{\infty} \left(|v|^{-3/4}-(1+v^2)^{-3/8}\right)\,dv>6.$$

Indeed the integrand is positive everywhere and its integral on $[-1,1]$ is greater than $2\int_0^1(v^{-3/4}-1)\,dv=6$. Also $c>11>(9/5)^4$. Therefore

$$I> -\frac{37}{4}+\frac{54}{5}=\frac{31}{20}. \tag{4}$$

Passing to the nearby circles

Inverse-branch integration and the change of scale give exactly

$$R_y^{1/4}\left(\ell(p_y)-\ell(z^4+p_y(0))\right) =\int_{\Gamma_y}\left(G_{F_y}(\sigma)-|\sigma|^{-3/4}\right)|d\sigma|. \tag{5}$$

The reference is correct because $|\sigma|^{-3/4}$ is the inverse-derivative density of $z^4$, and $\{z:z^4\in\Gamma_y\}$ has length $R_y^{1/4}\ell(z^4-C_y/R_y)$.

We justify convergence of the right side of (5) to $I$, including its singularities and expanding tail. Restrict $y$ to a small fixed compact interval about $Y$. Its critical points stay simple and separated and their second derivatives stay bounded away from zero. The local inverse function theorem in a square-root coordinate therefore gives uniform constants $A,\eta_0>0$ such that, near each critical value $f_y$,

$$G_{F_y}(\sigma)\le A|\sigma-f_y|^{-1/2}+A \quad (0<|\sigma-f_y|<\eta_0).$$

Since $\Gamma_y$ passes through $f_y$ and its radius tends to infinity, integrating this estimate over $|\sigma-f_y|<\eta$ gives $O(\eta^{1/2})$, uniformly for $y>Y$ sufficiently close to $Y$. One can see this directly from the chord formula $|\sigma-f_y|=2R_y|\sin(t/(2R_y))|$ in arclength $t$ from $f_y$. The reference has no singularity there: the distance from $0$ to $\Gamma_y$ is $|v_y|\to c>0$.

Away from these windows, smooth convergence of the circles and the implicit function theorem for simple roots give convergence on each bounded set. To control infinity uniformly, take $\rho^4=\sigma$ and $\omega^4=1$. For $F_y(z)=z^4+a_yz^2+b_yz$, its four inverse branches have the uniform expansions

$$z_\omega=\omega\rho-\frac{a_y}{4\omega\rho} -\frac{b_y}{4\omega^2\rho^2}+O(|\rho|^{-3}), \qquad \frac{dz_\omega}{d\sigma} =\frac{\omega}{4\rho^3} \left(1+\frac{a_y}{4\omega^2\rho^2} +\frac{b_y}{2\omega^3\rho^3}+O(|\rho|^{-4})\right).$$

These are convergent inverse series near infinity with a uniform radius for the bounded coefficient family; differentiation is legitimate there. On summing the moduli, the linear terms of orders $\rho^{-2}$ and $\rho^{-3}$ cancel because $\sum_{\omega^4=1}\omega^{-2}=\sum_{\omega^4=1}\omega^{-3}=0$. Consequently

$$\left|G_{F_y}(\sigma)-|\sigma|^{-3/4}\right| \le B|\sigma|^{-7/4}$$

beyond a fixed radius, with $B$ independent of $y$. The length of the part of any circle in a disc of radius $r$ is at most $2\pi r$ (and that of a line is at most $2r$). Summing over the annuli $2^jT<|\sigma|\le2^{j+1}T$ bounds the integral of $|\sigma|^{-7/4}$ on the tail by $4\pi T^{-3/4}\sum_{j\ge0}2^{-3j/4}$, independently of the circle. Thus the tails tend uniformly to zero.

First remove the critical windows and the tail, take $y\downarrow Y$ on the remaining compact arcs, and then let the windows shrink and the tail radius increase. This proves convergence in (5) to $I$. Inequality (4) proves the claimed limit and hence the strict length comparison for all $y>Y$ sufficiently close to $Y$.