It wouldn't work simply because there doesn't exist any entangled – or unentangled – state that would violate the uncertainty principle. The inequality
$$\Delta x_1\cdot \Delta p_1 \geq \frac\hbar 2$$
is a universal law of physics (which may be easily proved as a mathematical theorem if one assumes that the quantities are described by the maths of quantum mechanics) and it holds regardless of values or uncertainty or properties of $x_2,p_2$ (properties of another particle) or their entanglement with $x_1,p_1$ (subscripts $1,2$ label the two particles here).
So you can prepare a state in which $x_1$ is sharply determined and $p_2$ is sharply determined – they commute with each other – but because $x_1$ is sharply determined, it follows that $p_1$ is completely uncertain and similarly $x_2$ is completely uncertain, so the two particles just can't have the same positions or the same momenta. They're not entangled in the usual sense.
Alternatively, you may prepare entangled states in which the two particles always have the same position
$$ |\psi\rangle = \int_{-\infty}^{+\infty} dx \, f(x)\ |x\rangle \otimes |x+a\rangle $$
where the first tensor factor describes the first particle and the second tensor factor describes the other particle (I added a variable shift $a$ so that the particles don't overlap if you don't want to). But for a particle in this state, we can prove that the uncertainty $\Delta p_2$ is equal to the uncertainty $\Delta p_1$ simply because the state is totally symmetric with respect to the exchange of the two particles so the calculations of $\Delta p_2$ and $\Delta p_1$ are manifestly the same and yield the same results.
Because $\Delta p_1\geq \hbar / 2(\Delta x_1)$ by the uncertainty principle, it follows that – in this state –
$$\Delta p_2 \geq \frac{\hbar}{2\cdot \Delta x_1} $$
as well. The momentum of the second particle is greater than the usual multiple of the inverse uncertainty of the position of the first particle. This inequality doesn't hold for a general state because the two particles have nothing to do with each other. But if you impose an additional condition that the particles are entangled, then their properties are really the same and one may prove additional inequalities.
Your problem is that you seem to think that you are free to impose an arbitrary collection of conditions on the state vector and it still exists. But you're not free to do so. In other words, if you do so, your condition have no solution. Such states can't exist in Nature. They're mathematically impossible.
The uncertainty principle is a simple consequence of the idea that quantum mechanical operators do not necessarily commute.
In quantum mechanics, you find that the state which describes a state of definite value of an observable $A$ is not the state which describes a state of definite value for an observable $B$ if the commutator of both observables $[A,B]$ is not zero. (Formally, the two operators are not simultaneously diagonalizable.)
You just write down the definition of the standard deviation of the operator $A$ on a state $\psi$,
$$ \sigma_A(\psi) = \sqrt{\langle A^2\rangle_\psi - \langle A\rangle^2_\psi}$$
where $\langle\dot{}\rangle_\psi$ is the expectation value in the state $\psi$ and with a bit of algebraic manipulation (done e.g. on Wikipedia) we find that
$$ \sigma_A(\psi)\sigma_B(\psi)\ge \frac{1}{2}\lvert \langle[A,B]\rangle_\psi\lvert$$
Now, the standard deviation (or "uncertainty") of an observable on a state tells you how much the state "fluctuates" between different values of the observable. The standard deviation is, for instance, zero for eigenstates of the observable, since you always just measure the one eigenvalue that state has.
Plugging in the canonical commutation relation
$$ [x,p] = \mathrm{i}\hbar$$
yields the "famous" version of the uncertainty relation, namely
$$ \sigma_x\sigma_p\ge \frac{\hbar}{2}$$
but there is nothing special about position and momentum in this respect - every other operator pair likewise fulfills such an uncertainty relation.
It is, in my opinion, crucial to note that the uncertainty principle does not rely on any conception of "particles" or "waves". In particular, it also holds in finite-dimensional quantum systems (like a particle with spin that is somehow confined to a point) for observables like spin or angular momentum which have nothing to do with anything one might call "wavenature". The principle is just a consequence of the basic assumption of quantum mechanics that observables are well-modeled by operators on a Hilbert space.
The reason how "waves" enter is that the uncertainty relation for $x$ and $p$ is precisely that of the "widths" of functions in Fourier conjugate variables, and the Fourier relationship we are most familiar with is that between position and momentum space. That the canonical commutation relations are equivalent to such a description by Fourier conjugate variables is the content of the Stone-von Neumann theorem.
However, it is the description by commutation relations and not that by Fourier conjugacy that generalizes to all quantum states and all operators. Therefore, it is the commutation relation between the operators that should be seen as the origin of their quantum mechanical uncertainty relations.
Best Answer
Just look at the formal version of the Heisenberg uncertainty principle: $$ \sigma_x(\psi) \sigma_p(\psi) \geq \hbar/2,$$ where $\sigma_A(\psi) = \sqrt{\langle\psi\vert A^2\vert \psi\rangle - \langle \psi \vert A\rvert \psi\rangle^2}$ is the standard deviation of an operator $A$ for the state $\vert \psi \rangle$.
When we say a state has a "well-defined" or "precise" value of the observable $A$, we mean it is an eigenstate. It is straightforward to check that $\sigma_A(\psi) = 0$ in an eigenstate. So no state can be both an eigenstate of $x$ and of $p$, since that would mean $0\geq \hbar / 2$, which is clearly false.
The "simultaneously" means precisely that: A (time-dependent) state $\psi(t)$ may be an eigenstate of $x$ in one instance, and an eigenstate of $p$ in another, but it is impossible that $\psi(t_0)$ for any fixed $t_0$ is both.