On the Tchebyshev inequality ih the two-dimensional case

Let $\xi_1$, $\xi_2$ be indepent random variables satisfying the conditions
$$ \mathbf P\{\xi_1\ge0\}=\mathbf P\{\xi_2\ge0\}=1\quad\mathbf M\xi_1=\mathbf M\xi_2=1. $$
For positive $\Delta_1$ and $\Delta_2$, the inequality
\begin{gather*} \mathbf P\{\Delta_1\min(\xi_1,\xi_2)+\Delta_2\max(\xi_1,\xi_2)\ge c\}\le \\ \le\max[(\Delta_1+\Delta_2)^2,\Delta_2/(1-\Delta_1),2\Delta_2(1-\Delta_2)+\Delta_2^2] \end{gather*}
is proved. Moreover, if $\xi_1$ and $\xi_2$ are equally distributed, then it is proved that
$$ \mathbf P\{\Delta_1\min(\xi_1,\xi_2)+\Delta_2\max(\xi_1,\xi_2)\ge c\}\le\max[(\Delta_1+\Delta_2)^2;2\Delta_2(1-\Delta_2)+\Delta_2^2]. $$