<s>[INST] Determining all $(a,b)$ on the unit circle such that $2x+3y+1\le a(x+2)+b(y+3)$ for all $(x,y)$ in the unit disk In the middle of another problem, I came up with the following inequality which needed to be solve for $(a, b)$ : $$2x+3y+1\le a(x+2)+b(y+3)$$ for all $(x, y)\in\mathbb{R}^2$ with $x^2+y^2\le1.$
Here the solution for $(a, b)$ must be a subset of the unit circle, and I believe that it is a singleton.
Since I have no clue to solve it in this form, I tried to substitute some points of the closed unit disk and make a system of inequalities. But it seems like there should be a general way of solving these type of problems.
 [/INST] Rewrite $2x+3y+1\le a(x+2)+b(y+3)$ as $(2-a)x+(3-b)y\leq 2a+3b-1$. Solve for the intersection of the line $(2-a)x+(3-b)y=2a+3b-1$ with $x^2+y^2=1$ to get that the $x$-coordinates of the intersection are $$\frac{-2a^2-3ab+5a+6b-2\pm\sqrt{-(b-3)^2(3a^2+12ab+8b^2-12)}}{a^2-4a+b^2-6b+13}.$$ If this has two real solutions, then the inequality cannot be satisfied for all $(x,y)$ in the unit circle, so we need to find the intersection of $-(b-3)^2(3a^2+12ab+8b^2-12)\leq 0$ and $a^2+b^2\leq 1$. Simultaneously solving these equations, we find that the real solutions for $(a,b)$ are $(\frac{2}{\sqrt{13}},\frac{3}{\sqrt{13}})$ and $(-\frac{2}{\sqrt{13}},-\frac{3}{\sqrt{13}})$. It's easy to see that the first solution is valid and the second is invalid - plug in $(x,y)=(0,0)$, for instance.
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