On special type polynomial inequalities over integers

ADDED: As Mark Sapir and other are pointing out, if you only have $neq$'s, no $=$'s, $<$'s or $>$'s, then there is always a solution. That is to say, if $u_1$, $u_2$, ..., $u_N$ are nonzero polynomials, then there is always a lattice point where all the $u_i$ are nonzero. I assume you are asking the nontrivial question and allowing $<$'s and $>$'s:

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No. Any set of equations can be turned into a set of special equations. For example, if you have the equation $x^3 y^2 z + x^2 = 7$, just introduce new variables $x_1$, $x_2$, $x_3$, $y_1$, $y_2$ and $z_1$, and write down the special equations $x_1=x_2$, $x_2=x_3$, $y_1=y_2$ and $x_1 x_2 x_3 y_1 y_2 z + x_1 x_2 =7$. This is often called the polarization trick.

So special equations are no simpler than ordinary equations and, as I imagine you know, there is no algorithm to solve Diophantine equations.

I just noticed that you said "inequalities" not equalities. But any Diophantine equation can be rewritten as an inequality: $f(x,y,z)=0$ is the same as $-1 < f(x,y,z) < 1$, and any inequality as an equality: $z geq 0$ is equivalent to $exists (p,q,r,s) : z=p^2+q^2+r^2+s^2$. So this doesn't gain or lose you any generality.

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