Quadratic inequality with a variable consisting of more than one unknown The initial inequality was 
$$2^{2ax+1}+2^a ≤ 2^{ax}+2^{ax+a+1},$$
after substituting $2^{ax}$ with $y$ (so that $y>0$), and transferring all the terms from the RHS to the left I got 
$$2\cdot y^2 − y\cdot({2^{a+1} +1)} + 2^a ≤ 0.$$
I am stuck there, I tried finding the roots of its quadratic equation and the vertex of the function but nothing worked. So my question is how to find which values of $y$ satisfy the inequality for a and x being real parameters.

Setting $y=2^{ax}$ and a rearranging a bit yields
$$2y^2-y\leq 2^{a+1}y-2^a,$$
and both sides factor nicely, giving
$$y(2y-1)\leq2^a(2y-1).$$
Can you take it from here?
