Solution: Apply the **Cauchy-Schwarz Inequality** in the Engel form (also known as Titu’s Lemma):
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- All roots are non-negative, so valid. The sum of the roots is \( 0 + 4 + 9 = 13 \).
- Thus, the sum is \( \boxed{13} \).
- Question: Let \( x, y, z \) be positive real numbers such that \( x + y + z = 3 \). Find the minimum value of \( \frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x} \).
- \frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x} \geq \frac{(x + y + z)^2}{y + z + x} = \frac{3^2}{3} = 3
- Equality holds when \( \frac{x}{y} = \frac{y}{z} = \frac{z}{x} \), which implies \( x = y = z \). Given \( x + y + z = 3 \), we get \( x = y = z = 1 \). Substituting:
- \frac{1^2}{1} + \frac{1^2}{1} + \frac{1^2}{1} = 1 + 1 + 1 = 3
