Show that the least natural number a for which x + ax ^-2 > 2 for all x belonging to (0,infinity) is 2.

This can be simply solved using A.M > G.M

as a,x are positive(given conditions0

let the elements be : x/2 , x/2 , a/x²

so A.M = 1/3*[ x/2 + x/2 + a/x² ]

   G.M = [ x/2 * x/2 * a/x² ]^1/3

so its, x + a/x² > 3.(a/4)^1/3

     as x + a/x² > 2, it is clear that

        3.(a/4)^1/3   >= 2

          a >= 32/27

since a is natural number, min. possible is 2

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regards

Ramesh