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It´s imediact that \[ a \le cosA + \sqrt3senA \Leftrightarrow \frac {a}{2} \le cos(A - 60). \] Here we can think about angles of $ 60$ degrees... Maybe we should think about the midpoint of $ AB$, because of the $ a/2$. We can also note that two circles intersecting give us isosceles triangles. Midpoints and isosceles triangles make us think about perpendicular bisectors. We also have a lot of unitary lengths... These things indicate equilateral triangles. In fact, if we call $ E$ a point of interscection of the circles $ A$ and $ D$, we will have the equilateral triangle $ ADE$. Now, the extreme case must be when three of the circles concur. This gives us three unitary lengths... We also need to use that the triangle is acute-angled. All this makes us... Consider the circumcenter $ O$ of triangle $ ABD$. If it is covered by a circle centered in $ A$, $ B$ or $ D$ (hence covered by them all), all the triangle $ ABD$ is covered. It happens because the triangle is acute-angled, such that $ O$ is inside $ ABD$ and, being $ MNP$ its medial triangle ($ M$ opposite to $ A$, $ N$ to $ B$ and $ P$ to $ D$), $ MD < OD$, $ BP < OB$ and $ AN < AO$, then the circles intersect outside the triangle and cover it all. Obviously, $ ABD$ be covered implies $ CBD$ be covered. So the quadrilateral is covered $ \Leftrightarrow O$ is covered by the circle centered in, let´s say, $ D$. Let $ R$ the circumradius of $ ABD$. Now, we can prove that $ R < OC$. First, if we consider $ AD$ as a horizontal axis, the horizontal distance from $ O$ to $ D$ is $ ND$, that is less than the horizontal distance from $ C$ to $ M$, once $ \angle ADC > 90$. Of course, the perpendicular bisector from $ BD$, while goes from $ M$ to $ O$, approaches vertically the straight line through $ A$ and $ D$ (does somebody know how to put the $ \longleftrightarrow$ over $ AD$ in LaTeX?). Thus, $ O$ is also nearer to $ D$ that to $ C$ vertically, from which follows that $ R < OC$. Here we conclude that $ O$, and then the parallelogram, is (totally) covered if and only if $ R \le 1 \Leftrightarrow_{CL}^{SL} \frac {\sqrt {a^2 + 1 - 2acosA}}{sinA} \le 2 \Leftrightarrow a^2 + ( - 2cosA)a + 1 \le 4sin^2A \Leftrightarrow a^2 + ( - 2cosA)a + cos^2A - 3sin^2A \le 0 \Leftrightarrow (a - cosA + \sqrt3sinA)(a - cosA - \sqrt3sinA) \le 0.$ If $ a + \sqrt3sinA > cosA$, we can ignore the factor $ a + \sqrt3sinA - cosA$, obtaining what we wanted. Else, we have $ cosA > a$ and this also finishes the problem. Well, this solution is short, but not very elegant... Any suggestion to make it better?