Let $a>0$ and $\mathcal{F} = \{f:[0,1] \to \mathbb{R} : f \text{ is concave and } f(0)=1 \}.$ Determine $$\min_{f \in \mathcal{F}} \bigg\{ \left( \int_0^1 f(x)dx\right)^2 - (a+1) \int_0^1 x^{2a}f(x)dx \bigg\}.$$
1. **Given Conditions and Definitions:**
   - \( a > 0 \)
   - \( \mathcal{F} = \{ f : [0,1] \to \mathbb{R} \mid f \text{ is concave and } f(0) = 1 \} \)
   - We need to determine:
     \[
     \min_{f \in \mathcal{F}} \left\{ \left( \int_0^1 f(x) \, dx \right)^2 - (a+1) \int_0^1 x^{2a} f(x) \, dx \right\}
     \]

2. **Using Concavity and Boundary Condition:**
   - Since \( f \) is concave and \( f(0) = 1 \), for any \( x \in [0,1] \) and \( t \in [0,1] \), we have:
     \[
     f(tx + (1-t)0) \geq t f(x) + (1-t) f(0)
     \]
   - Specifically, for \( t = x^a \):
     \[
     f(x^{a+1}) \geq x^a f(x) + (1 - x^a) f(0)
     \]
   - Given \( f(0) = 1 \):
     \[
     f(x^{a+1}) \geq x^a f(x) + (1 - x^a)
     \]

3. **Multiplying by \((a+1)x^a\) and Integrating:**
   - Multiply both sides by \((a+1)x^a\):
     \[
     (a+1)x^{2a} f(x) + (a+1)(x^a - x^{2a}) \leq (a+1)x^a f(x^{a+1})
     \]
   - Integrate both sides from 0 to 1:
     \[
     (a+1) \int_0^1 x^{2a} f(x) \, dx + (a+1) \int_0^1 (x^a - x^{2a}) \, dx \leq (a+1) \int_0^1 x^a f(x^{a+1}) \, dx
     \]

4. **Evaluating the Integrals:**
   - Evaluate the integral \(\int_0^1 (x^a - x^{2a}) \, dx\):
     \[
     \int_0^1 x^a \, dx = \frac{1}{a+1}, \quad \int_0^1 x^{2a} \, dx = \frac{1}{2a+1}
     \]
     \[
     \int_0^1 (x^a - x^{2a}) \, dx = \frac{1}{a+1} - \frac{1}{2a+1} = \frac{2a+1 - (a+1)}{(a+1)(2a+1)} = \frac{a}{(a+1)(2a+1)}
     \]

5. **Substituting and Simplifying:**
   - Substitute back into the inequality:
     \[
     (a+1) \int_0^1 x^{2a} f(x) \, dx + \frac{a}{2a+1} \leq \int_0^1 f(x) \, dx
     \]

6. **Using the Fact \((\int_0^1 f(x) \, dx - \frac{1}{4})^2 \geq 0\):**
   - This implies:
     \[
     \int_0^1 f(x) \, dx \leq \left( \int_0^1 f(x) \, dx \right)^2 + \frac{1}{4}
     \]

7. **Combining Inequalities:**
   - Combine the inequalities:
     \[
     \left( \int_0^1 f(x) \, dx \right)^2 - (a+1) \int_0^1 x^{2a} f(x) \, dx \geq \frac{a}{2a+1} - \frac{1}{4}
     \]
     \[
     \left( \int_0^1 f(x) \, dx \right)^2 - (a+1) \int_0^1 x^{2a} f(x) \, dx \geq \frac{2a-1}{8a+4}
     \]

8. **Conclusion:**
   - The minimum value is:
     \[
     \boxed{\frac{2a-1}{8a+4}}
     \]