Solid angles of a tetrahedron

I think this statement does not hold. The following tetrahedron should give a counterexample:

$A=(-0.5, 0, 0)$, $B=(1,0,0)$, $C=(0,\varepsilon^2, \varepsilon)$, $D=(0,\varepsilon^2, -\varepsilon)$, $1>>\varepsilon>0$.

Here $BCD$ has largest area, but the spherical angles at $C$ and $D$ should be significantly larger than the one at $A$. (I have done the calculation only approximatively, but I believe it is true).

For $\epsilon=0.4$, the solid angles at $C$ and $D$ in Dmitri's example are more than twice the solid angle at $A$: $24.5^\circ$ vs. $12.1^\circ$. (And the area of $\triangle BCD = 0.41 > 0.32 = \triangle ABC$.)

            
Nice!

For random tetrahedra with vertices in the unit cube, in $100,000$ trials I get $8,355$ counterexamples including (after rounding) $\lbrace (0.57, 0.28, 0.95), (0.15, 0.54, 0.87), (0.45, 0.97, 0.68), (0.96, 0.38, 0.24) \rbrace$.

The reason I looked for a counterexample is that I don't think the statement is plausible. As you move around in the space of tetrahedra, the largest face changes. Also, the largest solid angle changes. There seems to be no general reason for these changes to happen at exactly the same place.

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Here is a construction based on the observation above. Let $A = (0,0,0)$, $B=(5,1,0)$, $C=(5,-2,0)$. These are the vertices of a triangle in the $xy$-plane. Let $D(t) = (0,0,t)$.

For small $t \gt 0$, $AD$ is small, so $\triangle DAB$ and $\triangle DAC$ have small areas. $\triangle DBC$ orthogonally projects onto $\triangle ABC$, so $\triangle DBC$ is slightly larger than $\triangle ABC$. When $t$ is large, $\triangle DAC$ becomes larger than $\triangle DBC$. The transition occurs at $t = \frac{3\sqrt{5}}{2} = 3.3541.$

The solid angle at $A$ is constant, while the solid angle at $B$ increases and eventually becomes larger. The transition occurs at $t = \frac{69+3\sqrt{29}}{20} = 4.2578$.

When $3.3541 \lt t \lt 4.2578$, the largest solid angle is at $A$, while the side with the largest area is $\triangle DAC$.