Prove that for all real $x > 0$ holds the inequality $$\sqrt{\frac{1}{3x+1}}+\sqrt{\frac{x}{x+3}}\ge 1.$$
For what values of $x$ does the equality hold?
To prove the inequality 
\[ \sqrt{\frac{1}{3x+1}} + \sqrt{\frac{x}{x+3}} \ge 1 \]
for all \( x > 0 \), we will use the method of substitution and algebraic manipulation.

1. **Substitution and Simplification:**
   Let \( y = \sqrt{\frac{1}{3x+1}} \). Then, we have:
   \[ y^2 = \frac{1}{3x+1} \]
   which implies:
   \[ 3x + 1 = \frac{1}{y^2} \]
   Solving for \( x \), we get:
   \[ x = \frac{1}{3y^2} - \frac{1}{3} \]

2. **Substitute \( x \) into the second term:**
   We need to express \( \sqrt{\frac{x}{x+3}} \) in terms of \( y \). Using the expression for \( x \):
   \[ \sqrt{\frac{x}{x+3}} = \sqrt{\frac{\frac{1}{3y^2} - \frac{1}{3}}{\frac{1}{3y^2} - \frac{1}{3} + 3}} \]
   Simplify the denominator:
   \[ x + 3 = \frac{1}{3y^2} - \frac{1}{3} + 3 = \frac{1}{3y^2} - \frac{1}{3} + \frac{9}{3} = \frac{1}{3y^2} + \frac{8}{3} \]
   Thus:
   \[ \sqrt{\frac{x}{x+3}} = \sqrt{\frac{\frac{1}{3y^2} - \frac{1}{3}}{\frac{1}{3y^2} + \frac{8}{3}}} = \sqrt{\frac{1 - y^2}{8y^2 + 1}} \]

3. **Combine the terms:**
   We need to show:
   \[ y + \sqrt{\frac{1 - y^2}{8y^2 + 1}} \ge 1 \]
   Let \( z = \sqrt{\frac{1 - y^2}{8y^2 + 1}} \). Then, we need to prove:
   \[ y + z \ge 1 \]

4. **Analyze the function:**
   Consider the function \( f(y) = y + \sqrt{\frac{1 - y^2}{8y^2 + 1}} \). We need to show that \( f(y) \ge 1 \) for \( y \in (0, 1) \).

5. **Check boundary conditions:**
   - When \( y = 0 \):
     \[ f(0) = 0 + \sqrt{\frac{1 - 0}{8 \cdot 0 + 1}} = 1 \]
   - When \( y \to 1 \):
     \[ f(1) = 1 + \sqrt{\frac{1 - 1}{8 \cdot 1 + 1}} = 1 \]

6. **Intermediate values:**
   For \( 0 < y < 1 \), we need to show that \( f(y) \ge 1 \). This can be done by analyzing the derivative of \( f(y) \) and showing that it is non-negative in the interval \( (0, 1) \).

7. **Equality condition:**
   The equality holds when \( y = 1 \), which corresponds to \( x = 1 \).

Therefore, the inequality is proven for all \( x > 0 \), and the equality holds when \( x = 1 \).

The final answer is \( \boxed{ x = 1 } \)