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.