5. Given real numbers $a, b$ satisfy $a>b>0, c$ satisfies that a triangle can be formed with $a, b, c$ as its side lengths. If the largest angle among the three interior angles of the triangle is to have the minimum value, then $c$ should be equal to ( ).
(A) $a$
(B) $\frac{a+b}{2}$
(C) $\sqrt{a b}$
(D) $b$
5. (A).

As shown in Figure 4, let the sides opposite to the three interior angles of $\triangle ABC$ be $a$, $b$, and $c$ respectively. Assuming points $A$ and $C$ are fixed, it is evident that vertex $B$ moves on the circumference of a circle centered at $C$ with radius $a$. Since in two triangles with two corresponding sides equal, the one with the larger included angle has the larger third side, side $c$ also increases as $B$ moves along the circumference. Clearly, as $c$ increases, $\angle A$ decreases, and $\angle A > \angle B$, thus $\max \{A, C\} = \max \{A, B, C\}$. Therefore, $c$ is the larger of $a$ and $b$. Since $a > b$, $c = a$ satisfies the condition.