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Let \(a\) and \(b\) be positive real numbers such that \(a + 2b = 1.\) Find the minimum value of
 \(\frac{2}{a} + \frac{1}{b}\)

 Aug 14, 2019
 #1
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+1

a = 1/2   and   b = 1/4

Minimum value of: 2/a  +  1/b =8

 Aug 15, 2019
 #2
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+3

Let a and b be positive real numbers such that \(a + 2b = 1\).  
Find the minimum value of  \(\dfrac{2}{a} + \dfrac{1}{b}\).

 

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{2}{a} + \dfrac{1}{b}} \\\\ &=& \dfrac{a+2b}{ab} \quad | \quad a + 2b = 1 \\\\ &=& \mathbf{\dfrac{1}{ab}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \large{GM} &\large{\leq}& \large{AM} \\\\ \sqrt{a(2b)} & \leq & \dfrac{ a+2b}{2} \quad | \quad a + 2b = 1 \\ \sqrt{a(2b)} & \leq & \dfrac{ 1}{2} \quad | \quad \text{square both sides} \\ 2ab & \leq & \dfrac{ 1}{4} \quad | \quad :2 \\ ab & \leq & \dfrac{ 1}{8} \\ && \boxed{ \text{If the sides of the inequality are either both positive or both negative, applies:} \\ \text{If the reciprocal value is formed on both sides of an inequality} \\ \text{the inequality sign turns around} } \\ \dfrac{1}{ab} & \geq & 8 \\ \hline \end{array}\)

 

The minimum value of  \(\mathbf{\dfrac{1}{ab}}=\dfrac{2}{a} + \dfrac{1}{b}\) is 8  

 

laugh

 Aug 15, 2019
edited by heureka  Aug 15, 2019

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