Explanation:

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.