\frac{a_1^2}{b_1} + \frac{a_2^2}{b_2} + \frac{a_3^2}{b_3} \ge \frac{(a_1 + a_2 + a_3)^2}{b_1 + b_2 + b_3}
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- a^3 + b^3 = 4^3 - 3 \cdot 3 \cdot 4 = 64 - 36 = \boxed{28}
- Question: Let $ x, y, z $ be positive real numbers such that $ x + y + z = 1 $. Find the minimum value of $ \frac{1}{x} + \frac{4}{y} + \frac{9}{z} $.
- Solution: We apply the **Cauchy-Schwarz inequality** in the Engel form (also known as Titu’s Lemma):
- But our expression is $ \frac{1}{x} + \frac{4}{y} + \frac{9}{z} = \frac{1^2}{x} + \frac{2^2}{y} + \frac{3^2}{z} $
- Apply Titu’s Lemma:
- \frac{1^2}{x} + \frac
