Let $a, b, c, d \ge 0$ such that $a^2 + b^2 + c^2 + d^2 = 4$. Prove that $$\frac{a + b + c + d}{2} \ge 1 +  \sqrt{abcd}$$ When does the equality hold?

Leonard Giugiuc and Valmir B. Krasniqi
Given \(a, b, c, d \ge 0\) such that \(a^2 + b^2 + c^2 + d^2 = 4\), we need to prove that 
\[ \frac{a + b + c + d}{2} \ge 1 + \sqrt{abcd} \]
and determine when equality holds.

1. **Substitution and Simplification:**
   Let \(a + b = m\), \(c + d = n\), \(ab = x\), and \(cd = y\). The inequality becomes:
   \[ \frac{m + n}{2} \ge 1 + \sqrt{xy} \]

2. **Convexity Argument:**
   Define the function:
   \[ f(x, y) = xy - \left(\frac{m + n - 2}{2}\right) \]
   We need to show that \(f(x, y)\) is convex in \(x\) and \(y\). The second partial derivatives are:
   \[ \frac{\partial^2 f(x, y)}{\partial x^2} = 0 \]
   \[ \frac{\partial^2 f(x, y)}{\partial y^2} = 0 \]
   Since the second partial derivatives are zero, \(f(x, y)\) is linear in \(x\) and \(y\), and thus convex.

3. **Boundary Values:**
   Since \(x = ab\) and \(y = cd\), we have \(0 \le x \le \frac{m^2}{4}\) and \(0 \le y \le \frac{n^2}{4}\). By the properties of convex functions, the maximum value of \(f(x, y)\) occurs at the boundaries:
   \[ \max f(x, y) = \max f(\alpha, \beta) \]
   where \(\alpha \in \{0, \frac{m^2}{4}\}\) and \(\beta \in \{0, \frac{n^2}{4}\}\).

4. **Case Analysis:**
   - **Case 1: \(x = \frac{m^2}{4}\) and \(y = \frac{n^2}{4}\):**
     This implies \(a = b\) and \(c = d\). We need to prove:
     \[ a + c \ge 1 + ac \]
     given \(a^2 + c^2 = 2\). Squaring both sides:
     \[ (a + c)^2 \ge (1 + ac)^2 \]
     \[ a^2 + 2ac + c^2 \ge 1 + 2ac + a^2c^2 \]
     Since \(a^2 + c^2 = 2\), we have:
     \[ 2 \ge 1 + a^2c^2 \]
     \[ 1 \ge a^2c^2 \]
     \[ (a^2 - 1)^2 \ge 0 \]
     which is always true.

   - **Case 2: \(x = 0\) and \(y = 0\) (i.e., \(abcd = 0\)):**
     This forces one of the variables to be zero. Assume \(d = 0\). We need to prove:
     \[ a + b + c \ge 2 \]
     given \(a^2 + b^2 + c^2 = 4\). Squaring both sides:
     \[ (a + b + c)^2 \ge 4 \]
     \[ a^2 + b^2 + c^2 + 2(ab + bc + ca) \ge 4 \]
     Since \(a^2 + b^2 + c^2 = 4\), we have:
     \[ 4 + 2(ab + bc + ca) \ge 4 \]
     \[ 2(ab + bc + ca) \ge 0 \]
     which is always true.

Thus, the inequality holds in both cases.

Equality holds when \(a = b = c = d = 1\), as this satisfies \(a^2 + b^2 + c^2 + d^2 = 4\) and \(\frac{a + b + c + d}{2} = 1 + \sqrt{abcd}\).

The final answer is \(\boxed{\frac{a + b + c + d}{2} \ge 1 + \sqrt{abcd}}\)