Then g' (x) = 2lnx/x.

According to Lagrange's mean value theorem,

ξ∈(a, b) exists, so

g(b)-g(b)=g'(ξ)(b-a)

g''(x)=2( 1-lnx)/x？

When x > e, g'' (x) < 0.

∴(e,e？ ), g'(x) monotonically decreases,

∴g'(ξ)>g'(e？ )=4/e？

∴ln？ b-ln？ a>4/e？ (b-a)