Math, asked by chandrima884, 1 year ago

if a,b,c are real numbers then find the least value of (a+b+c) (1/a+1/b+1/c))


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Answers

Answered by Anonymous
1

Heya....

Here is your answer...

You must be knowing that arithmetic mean greater than or equal to geometric mean.


Arithmetic mean of a,b,c is a+b+c / 3


Geometric mean of a,b,c is (abc)^1/3


This implies


a+b+c / 3 >= (abc)^1/3 ....1


Now reciprocate it so we get


1/3 (1/a +1/b+ 1/c) >= (1/a×1/b×1/c)^1/3 ....2


Multiply 1 and 2


So we get


1/9( a+b+c )(1/a +1/b+ 1/c) >= {(abc)^1/3 )}{(1/a×1/b×1/c)^1/3}


By cross multiplying we get,


( a+b+c )(1/a +1/b+ 1/c) >= 9


Thus, the minimum value of ( a+b+c )(1/a +1/b+ 1/c is 9.

Thanks...!!!

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Sorry baby 'wink'




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