Question 1164363
To prove the inequality $E|X+Y|^r \le 2^r(E|X|^r + E|Y|^r)$ for $r > 0$, we utilize the properties of absolute values and a specific power inequality.

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### **1. The Fundamental Inequality**
For any two real numbers $a$ and $b$ and any $r > 0$, the following inequality holds:
$$|a + b|^r \le [2 \max(|a|, |b|)]^r = 2^r \max(|a|^r, |b|^r)$$

Since $|a|^r$ and $|b|^r$ are both non-negative, we can bound the maximum of the two by their sum:
$$\max(|a|^r, |b|^r) \le |a|^r + |b|^r$$

Combining these, we get:
$$|a + b|^r \le 2^r (|a|^r + |b|^r)$$

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### **2. Application to Random Variables**
Let $a = X$ and $b = Y$. For every point in the sample space, the algebraic inequality holds:
$$|X + Y|^r \le 2^r (|X|^r + |Y|^r)$$

Now, we take the **Expected Value ($E$)** of both sides. Because the expectation is a linear operator and preserves inequalities (monotonicity), we have:
$$E[|X + Y|^r] \le E[2^r (|X|^r + |Y|^r)]$$

By the linearity of expectation, we can factor out the constant $2^r$ and split the sum:
$$E|X + Y|^r \le 2^r (E|X|^r + E|Y|^r)$$

$\blacksquare$

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### **3. Refinement for Different Values of $r$**
While the inequality above is always true for $r > 0$, it is actually a "loose" version of more specific inequalities:

* **Case $r \ge 1$ (Minkowski/Convexity):**
    Using the convexity of $\phi(t) = t^r$, we can derive a sharper bound:
    $$|X+Y|^r = 2^r \left| \frac{X+Y}{2} \right|^r \le 2^r \left( \frac{|X|^r + |Y|^r}{2} \right) = 2^{r-1}(|X|^r + |Y|^r)$$
* **Case $0 < r < 1$ (Subadditivity):**
    For $r$ in this range, the function is subadditive, meaning $|a+b|^r \le |a|^r + |b|^r$. In this case, the constant is simply $1$.

> **Summary:** Your requested inequality $E|X+Y|^r < 2^r(E|X|^r+E|Y|^r)$ is a universally valid (though conservative) bound for all $r > 0$ based on the maximum possible value the sum could take.