Call the first term on the right u, the second v, and note that uv is constant. This suggests using the geometricarithmetic mean inequality:
2u+vββ₯uvβ,
where u,v are any nonnegative numbers and equality holds iff u=v. Applying this inequality to our u and v gives
2f(x)ββ₯9β 4β, or f(x)β₯12
The value 12 is actually attained iff there is an x for which
9xsinx=xsinx4β, i.e., x2sin2x=94β.
Since x2sin2x is 0 when x=0 and it exceeds 1 when x=2Οβ, it follows that it equals the intermediate value 94β somewhere between 0 and 2Οβ. Thus the minimum value of f(x) is indeed 12β.