構(gòu)造函數(shù)法證明不等式的八種方法

1、1導(dǎo)數(shù)之構(gòu)造函數(shù)法證明不等式導(dǎo)數(shù)之構(gòu)造函數(shù)法證明不等式1、移項(xiàng)法構(gòu)造函數(shù)、移項(xiàng)法構(gòu)造函數(shù)【例1】1】已知函數(shù)，求證：當(dāng)時(shí)，恒有xxxf???)1ln()(1??xxxx?????)1ln(111【解】1111)(???????xxxxf∴當(dāng)時(shí)，，即在上為增函數(shù)01???x0)(??xf)(xf)01(??x當(dāng)時(shí)，，即在上為減函數(shù)0?x0)(??xf)(xf)0(???x故函數(shù)的單調(diào)遞增區(qū)間為，單調(diào)遞減區(qū)間()fx)01(?)0(??于是

2、函數(shù)在上的最大值為，因此，當(dāng)時(shí)，()fx)1(???0)0()(max??fxf1??x，即∴（右面得證），0)0()(??fxf0)1ln(???xxxx??)1ln(現(xiàn)證左面，令，111)1ln()(?????xxxg22)1()1(111)(???????xxxxxg則當(dāng)，0)()0(0)()01(?????????xgxxgx時(shí)當(dāng)時(shí)即在上為減函數(shù)，在上為增函數(shù)，)(xg)01(??x)0(???x故函數(shù)在上的最小值為，)(xg)

3、1(???0)0()(min??gxg∴當(dāng)時(shí)，，即1??x0)0()(??gxg0111)1ln(?????xx∴，綜上可知，當(dāng)111)1ln(????xxxxxx???????)1ln(1111有時(shí)2、作差法構(gòu)造函數(shù)證明、作差法構(gòu)造函數(shù)證明【例2】已知函數(shù)求證：在區(qū)間上，函數(shù)的圖象在函數(shù).ln21)(2xxxf??)1(??)(xf的圖象的下方；332)(xxg?【解】設(shè)，即，)()()(xfxgxF??xxxxFln2132)(23

4、???則=xxxxF12)(2????xxxx)12)(1(2???3記ｇ（ｘ）＝ｘｅｘ，則ｇ′(ｘ)＝ｅｘ－ｘｅｘ=(1x)ex，當(dāng)ｘ＞１時(shí)，ｇ′（ｘ）＜０，當(dāng)ｘ＜１時(shí)，ｇ′（ｘ）＞０知ｇ（ｘ）在(∞1)上為增函數(shù)在(1∞)上為減函數(shù)∴g(x)在x=1時(shí)取得最大值，即g(x)max=g(1)=1e∴a≥1e即a的取值范圍是[1e∞)(2)記F(X)=f(x)－(1x)=)0(1212????xxxex則F′(x)=ex1x令h(x)=

5、F′(x)=ex1x則h′(x)=ex1當(dāng)x0時(shí)h′(x)0∴h(x)在(0∞)上為增函數(shù)又h(x)在x=0處連續(xù)∴h(x)h(0)=0即F′(x)0∴F(x)在(0∞)上為增函數(shù)又F(x)在x=0處連續(xù)∴F(x)F(0)=0即f(x)1x6.6.對(duì)數(shù)法構(gòu)造函數(shù)（選用于冪指數(shù)函數(shù)不等式）對(duì)數(shù)法構(gòu)造函數(shù)（選用于冪指數(shù)函數(shù)不等式）例：證明當(dāng)2111)1(0xxexx?????時(shí)7.7.構(gòu)造形似函數(shù)構(gòu)造形似函數(shù)例：證明當(dāng)abbaeab???證