F(x)≤kx, x∈[ 1/2, 1] is equivalent to k≥m/x? +2/x+ 1，x∈[ 1/2， 1].

Let t = 1/x ∈ [1, 2], then k≥mt? +2t+ 1=m(t+ 1/m)？ + 1- 1/m=g(t)，t∈[ 1，2]。

K≥g(t)min。

Because it's m

When-1/ m

Therefore, when -2/3 ≤ m < 0, k ≥ m+3; When m & lt-2/3, k≥4m+5.