Constructor y=lnx/x

y'=[(lnx)'x-x'lnx]/x？ =( 1-lnx)/x？

When x> Iraq, Iraq

So f(x)=lnx/x is a decreasing function on (e, +∞).

x+a & gt； a & gte

So f (x+a) < f(a)

That is ln (x+a)/(x+a) < lna/a.

So AlN (x+a) < (a+x)lna.