The sides $a,b,c$ of triangle $ABC$ satisfy $a\le b\le c$. The circumradius and inradius of triangle $ABC$ are $R$ and $r$ respectively. Let $f=a+b-2R-2r$. Determine the sign of $f$ by the measure of angle $C$.
1. **Using Carnot's Theorem**: According to Carnot's theorem, we have:
   \[
   R + r = R (\cos A + \cos B + \cos C)
   \]
   This can be rearranged to:
   \[
   R + r = R (\cos A + \cos B + \cos C)
   \]
   Simplifying, we get:
   \[
   R + r = R \cos A + R \cos B + R \cos C
   \]

2. **Expressing \(a\) and \(b\) using the sine law**: By the sine law, we have:
   \[
   a = 2R \sin A \quad \text{and} \quad b = 2R \sin B
   \]

3. **Defining \(f\) and \(S\)**: Let:
   \[
   f = a + b - 2R - 2r
   \]
   We can rewrite \(f\) in terms of \(S\):
   \[
   S = \frac{a + b - 2R - 2r}{2R}
   \]
   It suffices to find the sign of \(S\) by the measure of angle \(C\).

4. **Substituting \(a\) and \(b\) into \(S\)**: Using the sine law expressions for \(a\) and \(b\), we get:
   \[
   S = \frac{2R \sin A + 2R \sin B - 2R - 2r}{2R}
   \]
   Simplifying, we have:
   \[
   S = \sin A + \sin B - 1 - \frac{r}{R}
   \]

5. **Using Carnot's theorem**: From Carnot's theorem, we know:
   \[
   R + r = R (\cos A + \cos B + \cos C)
   \]
   Therefore:
   \[
   \frac{r}{R} = \cos A + \cos B + \cos C - 1
   \]

6. **Substituting \(\frac{r}{R}\) into \(S\)**: We substitute \(\frac{r}{R}\) into the expression for \(S\):
   \[
   S = \sin A + \sin B - 1 - (\cos A + \cos B + \cos C - 1)
   \]
   Simplifying, we get:
   \[
   S = \sin A + \sin B - \cos A - \cos B - \cos C
   \]

7. **Expressing \(\cos C\)**: Note that:
   \[
   \cos C = -\cos (A + B) = \sin A \sin B - \cos A \cos B
   \]
   Therefore:
   \[
   S = \sin A + \sin B - \cos A - \cos B - (\sin A \sin B - \cos A \cos B)
   \]

8. **Simplifying \(S\)**: We simplify the expression for \(S\):
   \[
   S = \sin A + \sin B - \cos A - \cos B - \sin A \sin B + \cos A \cos B
   \]

9. **Defining \(g(x)\)**: Let:
   \[
   g(x) = \frac{1 - \sin x}{1 - \cos x}
   \]
   and let \(y = 90^\circ - A\). Then:
   \[
   S = (g(y) - g(B))(1 - \cos B)(1 - \cos y)
   \]

10. **Analyzing the sign of \(S\)**: Since \(1 - \cos B > 0\) and \(1 - \cos C > 0\) (since \(b\) is not the largest length and \(A > 0\)), \(S\) has the same sign as \(g(y) - g(B)\).

11. **Monotonicity of \(g(x)\)**: Note that in the interval \((0, 90^\circ)\), \(\sin x\) is an increasing function and \(\cos x\) is a decreasing function, hence \(g(x)\) is a decreasing function. Therefore:
    \[
    f > 0 \Leftrightarrow 90^\circ - A < B \Leftrightarrow C < 90^\circ
    \]
    \[
    f = 0 \Leftrightarrow 90^\circ - A = B \Leftrightarrow C = 90^\circ
    \]
    \[
    f < 0 \Leftrightarrow 90^\circ - A > B \Leftrightarrow C > 90^\circ
    \]

The final answer is \( \boxed{ f > 0 \Leftrightarrow C < 90^\circ } \), \(f = 0 \Leftrightarrow C = 90^\circ\), \(f < 0 \Leftrightarrow C > 90^\circ\).