if a,b,c are real numbers then find the least value of (a+b+c) (1/a+1/b+1/c))
Sanna113:
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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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