This proof is similar to River Li's, but it requires no machine calculation. Also, we will show more, namely that the original polynomial has three real roots, the smallest of which is at most
$$d:=\frac{5-\sqrt{15}}{10}=0.11270166\dotsc$$
This estimate is sharp, because in case of $\alpha=\beta=\gamma=\pi/2$, the polynomial is
$$20t^3-30t^2+12t-1=20\left(t-\frac{1}{2}\right)\left(t-\frac{5-\sqrt{15}}{10}\right)\left(t-\frac{5+\sqrt{15}}{10}\right).$$
We turn to the proof. 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
$$Q^3\geq R^2>\left(3\sqrt{\frac{3}{5}}-\frac{3}{2}\right)^2 Q^2\geq\frac{1}{2}Q^2\geq Q^3.$$