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a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2
We already know $ a^2 + b^2 = 1 $ and $ ab = \frac{1}{2} $, so
a^2b^2 = (ab)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}
Substituting into the identity:
a^4 + b^4 = 1^2 - 2 \cdot \frac{1}{4} = 1 - \frac{1}{2} = \frac{1}{2}
|a^4 + b^4| = \left| \frac{1}{2} \right| = \boxed{\frac{1}{2}}
Solution: To minimize
under the constraint $ x + y + z = 1 $, we apply the **Cauchy-Schwarz Inequality** in the following form:
Equality occurs when $ \frac{x}{y} = \frac{y}{z} = \frac{z}{x} $, which implies $ x = y = z $. Since $ x + y + z = 1 $, we have $ x = y = z = \frac{1}{3} $.
\frac{(1/3)^2}{1/3} + \frac{(1/3)^2}{1/3} + \frac{(1/3)^2}{1/3} = 3 \cdot \frac{1/9}{1/3} = 3 \cdot \frac{1}{3} = 1