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.
2026-03-29 20:15:27.1774815327
Quadratic inequality with a variable consisting of more than one unknown
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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?