If for any positive real number \ (X \) , are \ ({\ ln} xa \ mathrm {e} x-b + 1 \ leqslant 0 \) established \ ((\ mathrm {e} \) is the natural logarithm base number \ () \) , then \ (a b \) + minimum is \ (\ underline {\ qquad \ qquad}. \)
Analytical
note in question is given in the left side of the inequality \ (F (X) \) . Clearly \ (A> 0 \) , otherwise, there is always \ (x = \ mathrm {e } ^ b \) such that \ [F ( x) = ba \ mathrm {e } x-b + 1> 0. \] does not match the title set. Therefore, when \ (a> 0 \) we have \ [f (x) \ leqslant f \ left (\ dfrac { 1} {a \ mathrm {e }} \ right) = - {\ ln} a-1-b \ leqslant 0. \] so \ (B \ geqslant - {\ LN}. 1-A \) . thus \ [ a + b \ geqslant a-1 - {\ ln} a \ geqslant 0. \] Thus if and only if (\ (a, b) = (1, -1)) \ time, \ (A + B \) get the minimum \ (0 \) .