A problem by shanmuga rajesh

By applying A.M-G.M inequality on a,b,c &1/a,1/b,1/c

We get a+b+c/3 ≥ (abc)^1/3 ---->(1)

and

1/a+1/b+1/c / 3 ≥ 1/(abc)^1/3 ------>(2)

(1) x (2) =>[a+b+c][1/a+1/b+1/c] / 3 ≥ (abc)^1/3(1/abc)^1/3

=>[a+b+c][1/a+1/b+1/c] / 9 ≥ 1

=>[a+b+c][1/a+1/b+1/c] ≥ 9

therefore the least value of [a+b+c][1/a+1/b+1/c] is 9

by AM GM inequality

a+b+c/3≥cubic root of [abc]←equation 1

by AM GM inequality

1/a+1/b+1/c/3≥cubic root of 1/abc←equation 2

equation 1*2

[a+b+c][1/a+1/b+1/c]/9≥cubic root of [abc][1/abc]

[a+b+c][1/a+1/b+1/c]≥9*cubic root of 1

therefore [a+b+c][1/a+1/b+1/c]≥!9