How would one proceed with proving that there exists none $(x,y,z)$ such that all the following hold :
$$x+2y+7z = 0$$ $$\mathbb{P}(x + 3y + 9z \geq 0) = 1$$ $$\mathbb{P}(x + y + 5z \geq 0) = 1$$ $$\mathbb{P}(x + 13y + 10z \geq 0) = 1$$ $$\mathbb{P}(x + 3y + 9z >0 ) >0$$ $$\mathbb{P}(x + y + 5z > 0) >0 $$ $$\mathbb{P}(x + 13y + 10z > 0) >0$$
This is a "tool" that I need as a step to prove a case in a Mathematical Finance exercise.
Any ideas will be really appreciated.
$x=y=z=0$ is a solution to the first 4 constraints That means that there exists $a,b,c\ge 0$ such that $$\begin{align}x+2y+7z&=0\\x+3y+9z&=a\\x+y+5z&=b\\x+13y+10z&=c\end{align}$$ You must prove that the equations are not linearly dependent, so the only solution is $x=y=z=0$ for $a=b=c=0$. But then the last three inequalities are not obeyed (strict inequality)