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Home » \frac1n(n+2) = \fracAn + \fracBn+2. - Dachbleche24

\frac1n(n+2) = \fracAn + \fracBn+2. - Dachbleche24

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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\frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x}.
Solution: We are to minimize the expression
subject to $ x + y + z = 1 $, with $ x, y, z > 0 $.
Apply the **Cauchy-Schwarz Inequality** in the following form:
\left( \frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x} \right)(y + z + x...
Since $ x + y + z = 1 $, we have:
\left( \frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x} \right)(1) \geq 1...
\frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x} \geq 1
Equality holds when $ \frac{x}{y} = \frac{y}{z} = \frac{z}{x} $, which...
Check the expression at $ x = y = z = \frac{1}{3} $:

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