Prove $a\ge b \ge c \ge d \ge 1 \implies x^{4} -ax^{3}-bx^{2}-cx-d $ has no integer root. If $a,b,c,d$ are positive integers such that $a\ge b\ge c \ge d\ge 1$, prove that
$$ x^{4} -ax^{3}-bx^{2}-cx-d $$
has no integer root.

Attempt:
If there is an integer solution $x=x_{0}$, then
$$x_{0}^{4} = a x_{0}^{3}+ b x_{0}^{2}+c x_{0}+d$$
Notice that it is clear $x_{0} | d$.
Notice that $x_{0} \ne 0$ because $d>0$. If $x_{0}=1$ then, $1=a+b+c+d$, which is impossible because $a+b+c+d \ge 4$. So $x_{0} \notin \{0,1\}$. If $x_{0} = -1$, then $1 = (b-a) + (d-c) \le 0$, contradiction. So $x_{0} \notin \{-1,0,1\}$. So far we can say that $|x_{0}| \ge 2$.
Now if $x_{0} < 0$, then $x_{0}^{4} \ge 16 >  0$. But $a x_{0}^{3},cx_{0} < -1$  with
$$ |a x_{0}^{3} | > b x_{0}^{2} $$
$$ |c x_{0} | > d $$
so we have $x_{0}^{4} = ax_{0}^{3} + bx_{0}^{2} + cx_{0} + d < 0$ contradiction.
So we must have $x_{0} \ge 2$. Now, since $x_{0}| d$ then $d = e x_{0}$, where $e$ positive integer.
But this means $ a \ge b \ge c \ge d \ge x_{0}$, which means
$$ax_{0}^{3}+bx_{0}^{2}+cx_{0}+d > x_{0}^{4}$$
contradiction.
Some parts of this proof are not necessary I know, I was just working on it while writing it in this post.
Are the better/more elegant solutions?

The case $x=0$ is trivial.
For $x>0$, you have $x^4=(ax+b)x^2+(cx+d)$, with $ax+b\ge cx+d>1$. Let's say that $A=ax+b$ and $C=cx+d$. That means that $x^4=Ax^2+C$ and $C$ must be divisible by $x^2$ but $A\ge C\ge x^2$ so you can't find a solution.
For $x<0$, we have $A\le C\le0$ so $Ax^2+C$ will never be positive. So once again, no solution.
