If x, y, and z are positive numbers and x + y + z = 1, then the least value of the expression:
1/x + 1/y + 1/z is:
9
Solution
To minimize \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\) subject to \(x + y + z = 1\) where \(x, y, z > 0\), we use the AM-HM inequality.
Step 1: AM-HM Inequality
The AM-HM inequality states:
\[ \frac{x + y + z}{3} \geq \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \]
Since \(x + y + z = 1\), substituting into the inequality gives:
\[ \frac{1}{3} \geq \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \]
Step 2: Solve for \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\)
Rearranging the inequality:
\[ \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \leq \frac{1}{3} \]
Taking reciprocals (and reversing the inequality):
\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq 9 \]
Step 3: Equality Condition
Equality holds when \(x = y = z\). Since \(x + y + z = 1\), we get:
\[ x = y = z = \frac{1}{3} \]
Substituting into \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\):
\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{\frac{1}{3}} + \frac{1}{\frac{1}{3}} + \frac{1}{\frac{1}{3}} = 3 + 3 + 3 = 9 \]
Final Answer
The least value of \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\) is:
\[ \boxed{9} \]
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