The first question is simple. It shouldn't be difficult to take the derivative first. It can be concluded that when x = a (a > 0), f(x) takes the minimum value to calculate a= 1.

The second question Through the first question, we can understand that when a= 1, LNX+11is always greater than or equal to 0, so lnx- 1 is always greater than or equal to-1/x, then the formula we want to prove becomes.

Simple deformation formation, (xex-sinx)/x > 0 can be concluded, as long as xex-sinx > 0, it can be proved.

Then I won't say much about derivatives. Get the derivative function: ex( 1+x)-cosx Obviously, ex( 1+x) is always greater than 1, and -cosx is in the range of-1 to 1, so the derivative function is always greater than zero. Therefore, it is proved that.