Proving a complicated inequality involving integers

Let $a,b,c,d$ be integers such that $$\left( \begin{matrix} a & b \\ c & d \end{matrix} \right) = \left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right) \mod 2$$ $$ ad-bc =1$$ $$\left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \neq \left( \begin{matrix} \pm 1 & 0 \\ 0 & \pm 1 \end{matrix}\right) .$$

Let $(x,y)\in \mathbf{R}^2$ such that $-1\leq x\leq 1$ and $1/2 \leq y\leq 2$.

I highly suspect that $$c^2(y^2+2x^2) + a^2+d^2+2cx(d-a) + \frac{1}{y^2}(b-(d-a+3cx)x)^2 \geq 3.$$

The proof is actually very easy and was obtained after Parsa's answer.

In the case where $a$ and $d$ have the same sign, rewrite your expression as $$ c^2(y^2 + x^2) + (cx + (d-a))^2 + 2ad + \frac{1}{y^2}(b-(d-a+3cx)x)^2$$

Since each term is non-negative your inequality follows easily. Perhaps you can adjust this expression so your inequality follows in the case where $a$ and $d$ have opposite signs.