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Home » Thus, the value of $ c_3 $ is: - Crosslake

Thus, the value of $ c_3 $ is: - Crosslake

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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To find the minimum value of \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z...
\frac{x + y + z}{3} \geq \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}...
Substitute \( x + y + z = 6 \):
\frac{6}{3} \geq \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}
\geq \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}
\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \frac{3}{2}
Equality holds when \( x = y = z \). Since \( x + y + z = 6 \), set \(...
\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{2} + \frac{1}{2} +...
Thus, the minimum value is \(\boxed{\frac{3}{2}}\).
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