For positive real numbers $a_1, a_2, b_1, b_2$ satisfying $a_1 > a_2, b_1 > b_2$ holds $$ \begin{align} a_1+a_2 &> b_1+b_2 \tag{1}\label{eq1} \\ a_1a_2 &< b_1b_2 \tag{2}\label{eq2} \end{align} $$
Prove that then necessarily $a_1 > b_1 > b_2 > a_2$.
This is what I have tried
Introduction. Since $a_1 > a_2, b_1 > b_2$ then the largest number from $a_1, a_2, b_1, b_2$ can be $a_1$ or $b_1$. Also the smallest number can be $a_2$ or $b_2$.
1.) If $a_1 = b_1$ then it follows from $(1)$ that $a_2 > b_2$. However, then $(2)$ doesn't hold. $a_1 \neq b_1$.
2.) If $a_1 < b_1$. Let $a_1 = x - \varepsilon,\, b_1 = x + \varepsilon,\, \varepsilon > 0$. If we substitute this into $(1)$ we get $a_2 > b_2 + 2\varepsilon \implies a_2 > b_2$. From discussion in the introduction and from $a_1 > a_2$ it then follows $b_1>a_1>a_2>b_2$.
3.) If $a_1 > b_1$ then for $(2)$ to hold it must necessarily $a_2 < b_2$ and $b_1 > a_2$. So it follows that $a_1>b_1>b_2>a_2$. For example for $a_1 = 5,\, a_2 = 1,\, b_1 = 3,\, b_2 = 2$ both inequalities $(1)$ and $(2)$ hold.
As you can see in 2.) I haven't found some contradiction that would show that $b_1>a_1>a_2>b_2$ is not possible. Can someone help me with this?