I can prove that the following problem is undecidable: To determine, given two homogenous polynomials $p_1$ and $p_2$, whether or not the inequality $p_1\le p_2$ holds for all integer arguments.

Indeed, suppose there was an algorithm $A$ to determine whether $p_1\le p_2$ always holds. Then we can use this algorithm to determine whether any polynomial $f$ has an integer zero.

To see this, suppose $f=f(x_1,\ldots,,x_n)$ has total degree $d$. Let $$g(x_1,\ldots,,x_n,z)=z^d f(x_1/z,\ldots,x_n/z),$$ so $g$ is homogeneous of degree $d$.

I claim that $f$ has no integer zero if and only if the inequality
$$2z^d\le g(x_1,\ldots,,x_n,z)^2+z^{2d} $$
holds for all integer arguments. Note that the left and right hand sides are homogeneous polynomials, so if the claim is true then we can we can use algorithm $A$ to decide whether or not $f$ has an integer zero.

To verify the claim, suppose first that $f$ has no integer zero. If $z=1$ then the inequality reduces to $2\le g(x_1,\ldots,,x_n,1)^2+1$, i.e.,
$1\le f(x_1,\ldots,,x_n)^2$. If $z$ is different than 1, then already $2z^d\le z^{2d}$, therefore $2z^d\le g(x_1,\ldots,,x_n,z)^2+z^{2d}$. So if $f$ has no integer zero then the inequality holds for all integer arguments.

Conversely, if the inequality holds for all integer arguments, then put $z=1$ to obtain $1\le f(x_1,\ldots,,x_n)^2$.

What about the case that the coefficients of the $p_i$ are assumed to be positive? It would be interesting if in this case the problem was decidable.