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For a,b > 0, prove ab/(a+b) < a,b

 Feb 4, 2018
 #1
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ab / [ a + b ]   <  a   

 

Since a,b > 0. multiply both sides by a + b

 

ab <  a^2 + ab       subtract ab from  both sides

 

0 <  a^2           which is true

 

Likewise

 

ab / [ a + b ]  <  b

 

ab < ab + b^2

 

0 < b^2 

 

 

 

cool cool cool  

 Feb 4, 2018

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