# Solving a system of equations and inequalities

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.

• It looks like $(x,y,z)=(1,3,-1)$ satisfies all the conditions – WW1 Nov 13 '18 at 0:48
• @WW1 Thanks for the comment, there was actually a typo. Updated the question now ! – Rebellos Nov 13 '18 at 18:09

$$x=y=z=0$$ 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)
• Updated the question now, as the second per inequality equation is : $$x+y+5z$$ and the third one $$x+13y+10z$$ I apologize for that typo! Does the same approach still hold ? – Rebellos Nov 13 '18 at 18:11
• Yes. You must prove that you have solution only if $a\cdot b\cdot c=0$ – Andrei Nov 13 '18 at 18:19