Solution: Let $ A = rac{x^2 + y^2}{xy} $ and $ B = rac{xy}{x^2 + y^2} $. Then the expression becomes

February 22, 2026

$Solution: Let $ A = rac{x^2 + y^2}{xy} $ and $ B = rac{xy}{x^2 + y^2} $. Then the expression becomes$

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Solution: The divisors of 120 are $ 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 $. Those divisible by 5 are $ 5, 10, 15, 20, 30, 40, 60, 120 $. Summing them: $ 5 + 10 + 15 + 20 + 30 + 40 + 60 + 120 = 290 $. oxed{290}Question: Let $ x $ and $ y $ be positive real numbers representing the bandwidths (in Gbps) of two parallel data streams. Find the minimum value of
rac{x^2 + y^2}{xy} + rac{xy}{x^2 + y^2}
subject to $ x, y > 0 $.
A + B = rac{x^2 + y^2}{xy} + rac{xy}{x^2 + y^2}.
Let $ t = rac{x^2 + y^2}{xy} $. By the AM-GM inequality,
x^2 + y^2 \geq 2xy \Rightarrow rac{x^2 + y^2}{xy} \geq 2,

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