This proof is similar to River Li's, but it requires no machine calculation. Also, we will show a bit more, namely that the original polynomial has three real roots, the smallest of which is at most
$$d:=\frac{5-\sqrt{15}}{10}=0.11270166\dotsc$$
With the notation
$$Q:=\frac{\cos^2\alpha+\cos^2\beta+\cos^2\gamma}{3}
\qquad\text{and}\qquad
R:=\cos\alpha\cos\beta\cos\gamma,$$
the polynomial in the original post equals
$$p(t):=20t^3-30t^2+12(1-Q)t+(3Q-2R-1).$$
In particular,
$$p(1/2+v)=20v^3-3(1+4Q)v-(3Q+2R).$$
We claim that this cubic polynomial has a nonnegative discriminant, that is,
$$5(3Q+2R)^2\leq(1+4Q)^3.$$
To prove this inequality, we observe first that
$$5(3Q+2R)^2=5(9Q^2+12QR+4R^2)\leq 5(15Q^2+10R^2)=25(3Q^2+2R^2).$$
Using also that $R^2\leq Q^3$ holds by the AM-GM inequality, it suffices to show that
$$25(3Q^2+2Q^3)\leq (1+4Q)^3.$$
However, this one is clear, because
$$(1+4Q)^3-25(3Q^2+2Q^3)=(Q-1)^2(1+14Q)\geq 0.$$
We proved that the cubic polynomial $p(1/2+v)$ has a nonnegative discriminant, so it has three real roots (counted with multiplicity). So the original polynomial $p(t)$ has three real roots (counted with multiplicity), and the same is true of
$$p(d+s)=20s^3-6\sqrt{15}s^2+6(1-2Q)s+\left(6\sqrt{\frac{3}{5}}Q-3Q-2R\right).$$
It remains to show that this last polynomial has a nonpositive root. If this is not the case, then the polynomial has three positive roots, hence in particular
$$1-2Q>0>6\sqrt{\frac{3}{5}}Q-3Q-2R.\tag{$\ast$}$$
However, this is a contradiction, because it would yield that
$$Q^3\geq R^2>\left(3\sqrt{\frac{3}{5}}-\frac{3}{2}\right)^2 Q^2\geq\frac{1}{2}Q^2\geq Q^3.$$