Given the function f (x) = xlnx, if f (x) > = A-1 for all x > = 1, find the value range of real number a

analysis,
The minimum value of F (x) can be obtained by the method of derivation
f(x)=xlnx
Derivative f '(x) = LNX + 1
When x ≥ 1, f '(x) = LNX + 1 ≥ 1
Therefore, f (x) is an increasing function on [1, + ∞),
f(x)(mix)=f(1)=0
If A-1 ≤ 0, then f (x) ≥ A-1 must be constant,
∴a≦1.

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