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Grade: 11 

                        
Find the minimum value of (a+b+c)(1/a+1/b+1/c)? This is a question from progressions ,Algebra.
5 years ago

Answers : (1)

Anoopam Mishra
126 Points 
							
You must be knowing that AM \geqslant GM.
\frac{a+b+c}{3} \geqslant (abc)^{1/3} and \frac{1}{3} (\frac{1}{a} + \frac{1}{b} + \frac{1}{c} )\geqslant (\frac{1}{a} \frac{1}{b} \frac{1}{c})^{1/3}
Multiplying both these inequalities, we get
=> \frac{1}{9}(a+b+c) (\frac{1}{a} + \frac{1}{b} + \frac{1}{c} )\geqslant (abc)^{1/3} (\frac{1}{a} \frac{1}{b} \frac{1}{c})^{1/3}
=> (a+b+c) (\frac{1}{a} + \frac{1}{b} + \frac{1}{c} )\geqslant 9
The minimum value of (a+b+c) (\frac{1}{a} + \frac{1}{b} + \frac{1}{c} ) is 9.
5 years ago
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