不等式的证明例题

6.5函数的极值与最大值最小值1例1证xxxfxsine21)(2设xxxfxcose)(,0)(,10xfx.]1,0[)(上单调增加在所以xf定不出符号0)0(f且0)0(f且].1,0[)(Cxf0.21sine,102xxxx证明xxfxsine1)(6.4函数的单调性与曲线的凹凸性2)(,10xfx有时当0sine212xxx,10时当x,0)(,10xfx.]1,0[)(上单调增加在所以xf)(xf有)0(f.0].1,0[)(Cxf)0(f.0xxxfxcose)(即xxxfxsine21)(2上单调增加在]1,0[)(xf.21sine2xxx6.4函数的单调性与曲线的凹凸性3).(e4lnln,ee2222ababba证明设证法一,e4ln)(22xxx设则,e4ln2)(2xxx,ln12)(2xxx所以,e时当x,0)(x)(x故单调减少,从而,ee2时当x)e()(2x,ee2时即当x)(x单调增加.,ee2时当ba因此),()(ab即,e4lne4ln2222aabb故).(e4lnln222abab,0e4e422例26.4函数的单调性与曲线的凹凸性4).(e4lnln,ee2222ababba证明设2224()lnln().efxxaxa22ln4()exfxx21ln()20xfxx2()0fe()0fx()0fa()()0fbfa证法二6.4函数的单调性与曲线的凹凸性5).(e4lnln,ee2222ababba证明设证法三x2ln对函数,ln)(ttt,ln1)(2ttt,e时当t,0)(t所以)(t单调减少,从而).(e4lnln222abab在[a,b]上应用拉氏定理,得),(ln2lnln22abab.ba设则),e()(2即22eelnln即,e226.4函数的单调性与曲线的凹凸性6)1,,0,0(2)(21nyxyxyxyxnnnnttf)()(tf)(tf)]()([21yfxf即.2)(21nnnyxyx例3证,1nnt2)1(ntnn0yxt,0内任意两点对2yxf)0(t设图形是凹的.利用函数图形的凹凸性证明不等式:6.4函数的单调性与曲线的凹凸性7证法一用单调性证.法二用凹凸性证.,π2sin)(xxxf,π2cos)(xxf,0)0(f又,0)(xf因此.π2sinxx.π2sin,2π0:xxx时当证明不等式例4xxfsin)(设则,0,0)2π(f即所以f(x)的图形是凸的.6.4函数的单调性与曲线的凹凸性8证只要证.lnlnbaab令,lnln)(xaaxxfax则0)(afxaaxfln)(xa1,时当ax,)(时单调增加在axxf所以,时当ab)()(afbf即有,0lnlnbaabbaablnln得.abba,00也即.,eabbaab证明设例5ln()xfxx或者证明函数单调减少6.4函数的单调性与曲线的凹凸性9,0ba设证明不等式.1lnln222abababbaa证先证右边不等式.设axaxaxxlnln)(),0(ax0)(a)221(11)(xxaxaxxaxxax2)(20,时当ax)(x单调减少,故有)()(ax0即.lnlnaxaxaxbbb例66.4函数的单调性与曲线的凹凸性10,0ba设证明不等式.1lnln222abababbaa再证左边不等式.方法一设函数xxfln)(),0(ax由拉氏定理知,至少存在一点),,(ba使abablnln由于,0ba1从而.2lnln22baaababxx)(ln,1,222baab16.4函数的单调性与曲线的凹凸性11,0ba设证明不等式.1lnln222abababbaa再证左边不等式.方法二设)(2)ln)(ln()(22axaaxaxxf),0(ax因为axaxaxxxf21)()ln(ln2)(22xaxaxx2)()ln(ln20,时故当ax)(xf单调增加,故有)()(afxf00)(af即0)(2)ln)(ln(22axaaxax从而,0时当ab0)(2)ln)(ln(22abaabba即ababbaalnln2226.5函数的极值与最大值最小值12有对任意的试证明设,N,,,0nmax.)()(nmnmnmnmanmnmxax证设,)()(nmxaxxf11)()()(nmnmxanxxamxxf令0)(xf得nmmax分析求其在[0,a]上的最大值.,)()(nmxaxxf常数])([)(11nxxamxaxnm例7