= Solution
The substitution is nonsingular in the interior $x>0$, $0<y<1$, $0<z<x$. Its inverse is
$$
x=\sqrt{\alpha\beta},\qquad
y=\sqrt{\frac\beta\alpha},\qquad
z=\gamma\sqrt{\frac\alpha\beta}.
$$
Thus $y<1$ is equivalent to $\beta<\alpha$, while $z<x$ is equivalent to $\gamma<xy=\beta$. Positivity gives the full transformed domain
$$
\boxed{0<\gamma<\beta<\alpha<\infty.}
$$
The forward <Jacobian determinant> is
$$
\frac{\partial(\alpha,\beta,\gamma)}{\partial(x,y,z)}
=\det\begin{pmatrix}
1/y&-x/y^2&0\\
y&x&0\\
0&z&y
\end{pmatrix}
=2x>0.
$$
Therefore $x\,dx\,dy\,dz=\tfrac12\,d\alpha\,d\beta\,d\gamma$. The <boundary> faces where the formula fails have measure zero. The <change of variables formula> gives
$$
I=\frac12\int_0^\infty d\alpha\int_0^\alpha d\beta\int_0^\beta d\gamma\,
e^{-A\alpha-B\beta-C\gamma}.
$$
The integrand is nonnegative, so <Tonelli's theorem> permits reordering the <integration> even before convergence has been established. Integrating first in $\alpha$, then in $\beta$, gives
$$
\begin{aligned}
I
&=\frac12\int_0^\infty e^{-C\gamma}\,d\gamma
\int_\gamma^\infty e^{-B\beta}\,d\beta
\int_\beta^\infty e^{-A\alpha}\,d\alpha\\
&=\frac1{2A}\int_0^\infty e^{-C\gamma}\,d\gamma
\int_\gamma^\infty e^{-(A+B)\beta}\,d\beta\\
&=\frac1{2A(A+B)}
\int_0^\infty e^{-(A+B+C)\gamma}\,d\gamma\\
&=\boxed{\frac1{2A(A+B)(A+B+C)}}.
\end{aligned}
$$
The assumptions $A,B,C>0$ make all three integrals of <exponential functions> converge. Equivalently, the <ordered-cone substitution by products and ratios> can be followed by the increment variables $\alpha-\beta$, $\beta-\gamma$, $\gamma$, all independently positive and with <Jacobian determinant> one; the exponent then separates into three linear terms with coefficients $A$, $A+B$ and $A+B+C$.
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