Suppose that $a_1<1$, $a_3<1,$ $a_1+a_2+a_5>1$, $a_3+a_4+a_5>1,$ $a_1+a_2+a_3+a_4+a_5>2,$ and $b_1, b_2>0$.$a_1+a_2+a_3>1.$ For $x,y>0,$$x,y,z>0,$
(1) define a fucntion $$H(x,y)=\frac{x^{\frac{1}{2}}\int_0^{\infty}\int_0^{\infty}\frac{1}{t^{a_1}~ (1+t)^{a_2+1}~ z^{a_3}~(1+z)^{a_4}~ (1+t+z)^{a_5}}\exp\big\{-\frac{x}{1+t}-\frac{b_1 y}{1+t+z}\big\}dt dz}{\int_0^{\infty}\int_0^{\infty}\frac{1}{t^{a_1}~ (1+t)^{a_2}~ z^{a_3}~(1+z)^{a_4}~(1+t+z)^{a_5}}\exp\big\{-\frac{x}{1+t}-\frac{b_2 y}{1+t+z}\big\}dt dz}.$$$$H(x,y,z)=\frac{x^{\frac{1}{2}}\int_0^{\infty}\frac{1}{t^{a_1}~ (1+t)^{a_2+1}~ (1+t+z)^{a_3}}\exp\big\{-\frac{x}{1+t}-\frac{ y}{1+t+z}\big\}dt }{\int_0^{\infty}\frac{1}{t^{a_1}~ (1+t)^{a_2}~(1+t+z)^{a_3}}\exp\big\{-\frac{x}{1+t}-\frac{ y}{1+t+z}\big\}dt}.$$ Then $H(x,y)$$H(x,y,z)$ is uniformly bounded over $x,y$, i.e. there is a constant C, such that $H(x,y)\le C.$
(2) Furthermore, define $$L(x,y)=\frac{y^{\frac{1}{2}}\int_0^{\infty}\int_0^{\infty}\frac{1}{t^{a_1}~ (1+t)^{a_2}~ z^{a_3}~(1+z)^{a_4}~ (1+t+z)^{a_5+1}}\exp\big\{-\frac{x}{1+t}-\frac{b_1 y}{1+t+z}\big\}dt dz}{\int_0^{\infty}\int_0^{\infty}\frac{1}{t^{a_1}~ (1+t)^{a_2}~ z^{a_3}~(1+z)^{a_4}~(1+t+z)^{a_5}}\exp\big\{-\frac{x}{1+t}-\frac{b_2 y}{1+t+z}\big\}dt dz}.$$ Then$$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}.$$
Then $L(x,y)$ is also uniformly bounded over $x,y.$
How to prove it ??? Fedor Petrov gave the answer for the case when $b_1=b_2.$ I still didn't know how to prove the general case. I need your help.
Maybe, the following result is useful:
If $a_1<1$ and $a_1+a_2>1,$ then $$f(x)\equiv\int_0^{\infty}\frac{1}{t^{a_1}~ (1+t)^{a_2}}\exp\big\{-\frac{x}{1+t}\big\}dt\approx C_1\min\{C_2,x^{1-a_1-a_2}\}$$ for some positive constants $C_1$ and $C_2$.