Lebesgue's universal cover problem asks for a planar set of least possible area that contains a congruent copy of every planar set of diameter at most one. We work in the convex Brass$-$Sharifi three-test-set framework, where the test sets are a disk, an equilateral triangle, and a regular pentagon of diameter one. For each normalized placement $v$, let $A(v)$ denote the area of the convex hull of these three test sets. We construct a finite certificate proving $A(v) \ge 0.83201$ throughout the admissible normalized domain. The threshold $0.83201$ slightly improves the Brass$-$Sharifi lower bound $0.832$ within the same convex three-test-set framework. The proof is a finite-cover argument. The admissible domain is covered by finitely many parameter domains, and each domain carries a local lower-bound certificate. Most domains are handled by supporting local records. On the witness domains, the local bound is obtained from an inner-witness polygon construction. The witness points lie in the three test sets and determine an ordered polygonal region certified to be simple and positively oriented. Its area is bounded below by interval orientation and shoelace estimates. Since this certified polygonal region lies inside the corresponding convex hull, its area gives a lower bound for the hull area. Combining the local inequalities with the finite cover yields $α_{cvx} \ge 0.83201$, where $α_{cvx}$ is the infimum of the areas of convex universal covers.

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