# Inequality involving Weibull distribution

define R(x) as the weibull survival function

$$R_1(t)=e^{-\alpha_1 t^{\beta_1}}$$

$$R_2(t)=e^{-\alpha_2 t^{\beta_2}}$$

with $$\alpha_1, \alpha_2 > 0$$ and $$\beta_1, \beta_2 >1$$

$$\phi_1(t)=\int_{t}^{\infty}R_1(t) \mathrm{d}t$$

$$\phi_2(t)=\int_{t}^{\infty}R_2(t) \mathrm{d}t$$

I am trying to prove the following inequality (verified by matlab):

$$R_1(x)R_1(y)\phi_2(x+y)\phi_1(0)+R_2(x)R_2(y)\phi_1(x+y)\phi_2(0) \leq R_1(x)R_2(y)\phi_2(x)\phi_1(y)+R_2(x)R_1(y)\phi_1(x)\phi_2(y)$$

where $$x,y \geq 0$$

p.s. i found that for the function $$\phi$$, the following inequality holds: $$\phi(x)\phi(y)\geq \phi(x+y)\phi(0)$$ but i dont have other ideas...

• It seems something is wrong, because if $\alpha, x>0$ and $\beta>1$ then an integral $\int_{x}^\infty \alpha t^\beta dt$ diverges. – Alex Ravsky Nov 27 '18 at 18:42
• Thank you for your comments Alex Ravsky. In fact I forgot the exponential... – Xingheng Liu Dec 6 '18 at 19:27