<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] Since $(a,b)$ lies on the unit circle and $(a,b)$ satisfies the linear inequality, it must be tangent to the circle. So first find the tangent line to the unit circle at $(\cos t, \sin t)$. This is:
$$y - \sin t = (-\cot t)(x - \cos t) \implies y = -(\cot t)x + \csc t.$$
Now the equality case of the condition $2x+3y+1 = a(x+2)+b(y+3)$ rearranges to $(3-b)y = (a-2)x$ $ + (2a + 3b - 1)$. Using the identity $\csc^2 t - \cot^2 t = 1$ and comparing coefficients:
$$\frac{1}{(3-b)^2} \left((2a + 3b - 1)^2 - (a-2)^2 \right) = 1$$
$$(4a^2 + 12ab + 9b^2 - 4a - 6b + 1) - (a^2 - 4a + 4) = 9 - 6b + b^2$$
$$3a^2 + 12ab + 8b^2 - 12 = 0$$
and now substitute $12 = 12(a^2+b^2)$:
$$-9a^2 + 12ab - 4b^2 = 0$$
$$-(3a - 2b)^2 = 0 \implies b = 3a/2$$
thus $(3b/2)^2 + b^2 = 1 \implies a = \frac{2}{\sqrt{13}}, b = \frac{3}{\sqrt{13}}$. We discard the negative solutions as substituting $(x,y) = (0,0)$ into the condition we need $1 ≤ 2a + 3b$.
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