Suppose $a,b,c,d$ are real numbers greater than $1.$ Given that \begin{align*} a + b + c + d &= -x \\ ab + ac + ad + bc + bd + cd &= y \\ abc + abd + acd + bcd &= -z \\ abcd &= 858, \end{align*} prove that $x + y + z < 3abcd.$
I first noted that $x, z < 0$ and that the problem asked for the proof that $x + y + z < 2574,$ but I didn't know where to go from there.
From the $y$ and $z$ equations we get $$ab + ac + abc + abd + acd + bcd \gt y$$ $$ab + ac \gt y+z$$ as $a, b, c, d$ are greater than 1. We can then subtract the $x$ equation to get $$ab + ac - a - b - c - d \gt x + y + z $$ so $$ab + ac \gt x + y + z $$ as $a, b, c, d$ are positive so $$3abcd \gt 2abcd \gt ab + ac \gt x + y + z$$ I can't see any errors in my method, although I am not certain of it as it doesn't use the last equation