Question

已知函数f(x)=xlnx.

(1)若f(x)≥ax-1对任意x＞0恒成立,求实数a的取值范围;

(2)若a＞0,b＞0,证明f(a)+(a+b)ln2≥f(a+b)-f(b).

Answer 1

(1)解法一:由f(x)≥ax-1,得xlnx≥ax-1,

即ax≤xlnx+1,

∵x＞0,∴a≤lnx+在x＞0上恒成立.

令g(x)=lnx+,

由g′(x)==0,得x=1.

∵x＞1时,g′(x)＞0,0＜x＜1时,g′(x)＜0,

∴g(x)在(0,1)上为减函数,在(1,+∞)上为增函数.

∴g(x)_min=g(1)=1.∴a≤1.

解法二:令g(x)=xlnx-ax+1,则g′(x)=lnx+1-a,

由g′(x)=0,得x=e^a-1,

当x∈(0,e^a-1)时,g′(x)＜0,x∈(e^a-1,+∞)时,g′(x)＞0,

∴g(x)在(e^a-1,+∞)上为增函数,在(0,e^a-1)上为减函数.

∴g(x)_min=g(e^a-1)=e^a-1lne^a-1-ae^a-1+1

=(a-1)e^a-1-ae^a-1+1

=-e^a-1+1.

要使f(x)≥ax-1在x＞0上恒成立,

即使g(x)≥0在x＞0上恒成立,也即g(x)_min≥0恒成立,

由-e^a-1+1≥0,得e^a-1≤1,即a≤1.

(2)证明:令h(a)=f(a)+(a+b)ln2-f(a+b)+f(b)

=alna+(a+b)ln2-(a+b)ln(a+b)+blnb.

∵h′(a)=lna+1+ln2-1-ln(a+b)=ln,

当a＞b＞0时,h′(a)＞0;

当0＜a＜b时,h′(a)＜0,

∴h(a)在(0,b)上为减函数,在(b,+∞)上为增函数.

∴h(a)_min=h(b)=f(b)+2bln2-f(2b)+f(b)

=2f(b)+2bln2-f(2b)

=2blnb+2bln2-2bln2b=2bln=0.

∴h(a)≥0,即f(a)+(a+b)ln2≥f(a+b)-f(b).