Daily questions _191128

Known \ (\ FORALL X> 0, X \ mathrm {E}} ^ {2x -kx - {\ X-LN}. 1 \ geqslant 0 \) , realistic number \ (K \) . The range
resolution:
a method is equivalent to the original inequality \ [\ forall x> 0, k \ leqslant \ dfrac {1} {x} \ cdot \ left (x \ mathrm {e} ^ {2x} - {\ ln} x-1 \ right) = \ dfrac {1} {x} \ cdot \ left (\ mathrm {e} ^ {2x + {\ ln} x} -. {\ ln} x-1 \ right) \] we who are familiar \ ( \ forall x \ in \ mathbb { R}, \ mathrm {e} ^ x \ geqslant x + 1. \) so \ [RHS \ geqslant \ dfrac { 1} {x} \ cdot \ left (2x + {\ ln} x + 1 -. {\ ln } x-1 \ right) = 2 \] if and only if \ (2x + {\ ln} x = 0 \) above inequality is made equal sign, the apparent presence of \ (x_0 \ in \ left (\ dfrac {1} { \ mathrm {e}}, 1 \ right) \) meet the first type, the \ (K \) ranges \ ((- \ infty, 2] \) .
method two inequality separation parameter equivalent to the original\ [\ forall x> 0, g (x) = \ mathrm {e} ^ {2x} - \ dfrac {1} {x} - \ dfrac {{\} x ln} {x} \ geqslant k \.] derivative available \ [g '(x) = 2 \ mathrm {e} ^ {2x} + \ dfrac {{\ ln} x} {x ^ 2}, x> 0. \] derivative available again \ [g '' (x) = 4 \ mathrm {e} ^ {2x} + \ dfrac {1-2 {\ ln} x} {x ^ 3}, x> 0. \] so \ [\ forall x> 0, g '' (x) \ geqslant 4 + 0- \ dfrac 23 \ cdot \ dfrac {1} {\ mathrm {e}}> 0. \] so \ (g '(x) \ ) in the interval \ ( (0, + \ infty) \ ) monotonically increasing, and because \ [g '(1)> 0> g' \ left (\ dfrac {1} {\ mathrm {e} ^ 3} \ right). \] Therefore \ (g '(x) \ ) there must be a unique zero point \ (x_0 \ in \ left (\ dfrac {. 1} {\ mathrm {E} ^. 3},. 1 \ right) \) , that is \ [2x_0 ^ 2 \ mathrm {e} ^ { 2x_0} = -. {\ ln} x_0 \] provided \ (T = x_0 ^ 2 \ mathrm {E} ^ {2x_0} \) , then \ (2t = - {\ ln } x_0 \) , and\ [{\ ln} t = {\ ln} (x_0 ^ 2 \ mathrm {e} ^ {2x_0}) = 2 {\ ln} x_0 + 2x_0. \] so \ ({\ ln} t + 2t = { \ ln} x_0 + 2x_0, \ ) so \ (x_0 = T \) , so we have \ [\ forall x> 0, g (x) \ geqslant g (x_0) = \ dfrac {t} {x_0 ^ 2} - \ dfrac {1} {x_0
} + \ dfrac {2t} {x_0} = \ dfrac {1} {x_0} -. \ dfrac {1} {x_0} + \ dfrac {2x_0} {x_0} = 2 \] Thus \ (K \) ranges \ ((- \ infty, 2] \) .