To prove the given inequality, we can use the concept of convexity and Jensen's inequality. Let's start by breaking down the problem and then provide the proof.

Problem Restatement

You are given that:

$\int \min \left(\frac{P}{R_1} p_{X}(x), p_{\hat{X}_{1}}(x)\right) dx \geq P$ $\int \min \left(\frac{P}{R_2} p_{X}(x), p_{\hat{X}_{2}}(x)\right) dx \geq P$

You need to prove that for every $\gamma \in [0,1]$,

$\int \min \left(\frac{P}{\gamma R_1 + (1-\gamma) R_2} p_{X}(x), \gamma p_{\hat{X}_{1}}(x) + (1-\gamma)p_{\hat{X}_{2}}(x)\right) dx \geq P$

Proof

Define the following function for the integral: $F(R, p_{\hat{X}}) = \int \min \left(\frac{P}{R} p_X(x), p_{\hat{X}}(x)\right) dx$

You are given that: $F(R_1, p_{\hat{X}_1}) \geq P$ $F(R_2, p_{\hat{X}_2}) \geq P$

We need to prove: $F(\gamma R_1 + (1-\gamma) R_2, \gamma p_{\hat{X}_1} + (1-\gamma) p_{\hat{X}_2}) \geq P$

Now consider the convex combination of the two integrals: $F(\gamma R_1 + (1-\gamma) R_2, \gamma p_{\hat{X}_1} + (1-\gamma) p_{\hat{X}_2}) = \int \min \left(\frac{P}{\gamma R_1 + (1-\gamma) R_2} p_X(x), \gamma p_{\hat{X}_1}(x) + (1-\gamma) p_{\hat{X}_2}(x)\right) dx$

Step 1: Use the convexity of the minimum function

For any $a, b, c, d \geq 0$, the minimum function satisfies the following convexity property: $\min(\gamma a + (1-\gamma) b, \gamma c + (1-\gamma) d) \geq \gamma \min(a, c) + (1-\gamma) \min(b, d)$

Applying this property to our case, we get: $\min \left(\frac{P}{\gamma R_1 + (1-\gamma) R_2} p_X(x), \gamma p_{\hat{X}_1}(x) + (1-\gamma) p_{\hat{X}_2}(x)\right) \geq \gamma \min\left(\frac{P}{R_1} p_X(x), p_{\hat{X}_1}(x)\right) + (1-\gamma) \min\left(\frac{P}{R_2} p_X(x), p_{\hat{X}_2}(x)\right)$

Step 2: Integrate both sides

Integrating both sides with respect to $x$, we obtain: $\int \min \left(\frac{P}{\gamma R_1 + (1-\gamma) R_2} p_X(x), \gamma p_{\hat{X}_1}(x) + (1-\gamma) p_{\hat{X}_2}(x)\right) dx \geq \gamma \int \min\left(\frac{P}{R_1} p_X(x), p_{\hat{X}_1}(x)\right) dx + (1-\gamma) \int \min\left(\frac{P}{R_2} p_X(x), p_{\hat{X}_2}(x)\right) dx$

Step 3: Apply the given conditions

By the given conditions, we know: $\int \min\left(\frac{P}{R_1} p_X(x), p_{\hat{X}_1}(x)\right) dx \geq P$ $\int \min\left(\frac{P}{R_2} p_X(x), p_{\hat{X}_2}(x)\right) dx \geq P$

Thus, we have: $\gamma \int \min\left(\frac{P}{R_1} p_X(x), p_{\hat{X}_1}(x)\right) dx + (1-\gamma) \int \min\left(\frac{P}{R_2} p_X(x), p_{\hat{X}_2}(x)\right) dx \geq \gamma P + (1-\gamma) P = P$

Conclusion

Since the right-hand side is greater than or equal to $P$, we have: $\int \min \left(\frac{P}{\gamma R_1 + (1-\gamma) R_2} p_X(x), \gamma p_{\hat{X}_1}(x) + (1-\gamma) p_{\hat{X}_2}(x)\right) dx \geq P$

This completes the proof.

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Would you like further details on any of these steps or have any other questions?

Related Questions:

1. What is Jensen's inequality and how does it apply in probability theory?
2. How does the convexity of the minimum function help in inequalities?
3. Can you explain the role of probability density functions in this proof?
4. How does this proof change if $\gamma$ is fixed at a certain value?
5. What are some real-world applications of this type of inequality?

Tip:

When proving inequalities involving integrals and probability distributions, leveraging properties like convexity and Jensen's inequality often simplifies the process significantly.