Suppose that $a\_1<1$, $a\_1+a\_2+a\_3>1.$ For $x,y,z>0,$ 


(1) define a fucntion
$$H(x,y,z)=\frac{x^{\frac{1}{2}}\int\_0^{\infty}\frac{1}{t^{a\_1}~ (1+t)^{a\_2+1}~ 
 (1+t+z)^{a\_3}}\exp\big\{-\frac{x}{1+t}-\frac{ y}{1+t+z}\big\}dt }{\int\_0^{\infty}\frac{1}{t^{a\_1}~ (1+t)^{a\_2}~(1+t+z)^{a\_3}}\exp\big\{-\frac{x}{1+t}-\frac{ y}{1+t+z}\big\}dt}.$$
Then $H(x,y,z)$ is uniformly bounded over $x,y$, i.e. there is a constant C, such that
$H(x,y)\le C.$


(2) Furthermore, define 
$$L(y,z)=\frac{y^{\frac{1}{2}}\int\_0^{\infty}\frac{1}{t^{a\_1}~ (1+t)^{a\_2}~ 
 (1+t+z)^{a\_3+1}}\exp\big\{-\frac{ y}{1+t+z}\big\}dt }{\int\_0^{\infty}\frac{1}{t^{a\_1}~ (1+t)^{a\_2}~(1+t+z)^{a\_3}}\exp\big\{-\frac{y}{1+t+z}\big\}dt}.$$


Then $L(x,y)$ is also uniformly bounded over $x,y.$



Uniformly Bounded (updating)
We may assume that $x>100(1+|a\_2|+|a\_3|)^2$, for other $x$ simply $C=10(1+|a\_2|+|a\_3|)$ works perfectly.


Partition the integral in the numerator onto two parts: $I\_1$ over $(0,\sqrt{x}]$ and $I\_2$ over $[\sqrt{x},\infty)$. The second part does not exceed the denominator, since $\sqrt{x}(1+t)^{-1}<1$ for all $t\geqslant \sqrt{x}$. It suffices to estimate $I\_1\leqslant c I\_2$ for some fixed $c>0$ (depending only on $a\_1,a\_2,a\_3$). For this it suffices to prove the pointwise estimate $f(t)\leqslant 2^{a\_1} f(2t)$ for all $t\in (0,\sqrt{x})$ where $f$ is the integrated function. Indeed, integrating this over $(0,\sqrt{x})$ we get $\int\_0^{\sqrt{x}}f(t)dt\leqslant 2^{a\_1}\int\_0^{\sqrt{x}}f(2t)dt=2^{a\_1-1}\int\_0^{2\sqrt{x}} f(t)dt$, thus $I\_1(1-2^{a\_1-1})\leqslant 2^{a\_1-1}I\_2$ as desired. We have
$$
\frac{f(t)}{2^{a\_1}f(2t)}=\left(\frac{1+2t}{1+t}\right)^{a\_2+1}\cdot 
\left(\frac{1+2t+z}{1+t+z}\right)^{a\_3}\cdot \exp\left(-\frac{x}{1+2t}\left(\frac{1+2t}{1+t}-1\right)-
\frac{z}{1+2t+z}\left(\frac{1+2t+z}{1+t+z}-1\right)
\right).
$$
We estimate $\frac{1+2t+z}{1+t+z}\leqslant \frac{1+2t}{1+t}$ and use the estimate $1+s\leqslant e^s$ for $s:=\frac{1+2t}{1+t}-1$. Also estimate $-\frac{z}{1+2t+z}\left(\frac{1+2t+z}{1+t+z}-1\right)\leqslant 0$ in the exponent. We get
$$
\frac{f(t)}{2^{a\_1}f(2t)}\leqslant \exp\left(\left(1+|a\_2|+|a\_3|-\frac{x}{1+2t}\right)s\right)\leqslant 1,
$$
since $\frac{x}{1+2t}\geqslant \frac{x}{1+2\sqrt{x}}>\frac13\sqrt{x}>1+|a\_2|+|a\_3|$ due to our assumptions. 


