Home / \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}

\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}

February 22, 2026

$\frac{a_1^2}{b_1} + \frac{a_2^2}{b_2} + \frac{a_3^2}{b_3} \ge \frac{(a_1 + a_2 + a_3)^2}{b_1 + b_2 + b_3}$

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