PAGE1-微專題20一元不等式的證明利用函數(shù)性質(zhì)與最值證明一元不等式是導(dǎo)數(shù)綜合題常涉及的一類問(wèn)題，考察學(xué)生構(gòu)造函數(shù)選擇函數(shù)的能力，體現(xiàn)了函數(shù)最值的一個(gè)作用——每一個(gè)函數(shù)的最值帶來(lái)一個(gè)恒成立的不等式。此外所證明的不等式也有可能對(duì)后一問(wèn)的解決提供幫助，處于承上啟下的位置。一、基礎(chǔ)知識(shí)：1、證明方法的理論基礎(chǔ)（1）若要證SKIPIF1<0(SKIPIF1<0為常數(shù)）恒成立，則只需證明：SKIPIF1<0，進(jìn)而將不等式的證明轉(zhuǎn)化為求函數(shù)的最值（2）已知SKIPIF1<0的公共定義域?yàn)镾KIPIF1<0，若SKIPIF1<0，則SKIPIF1<0證明：對(duì)任意的SKIPIF1<0，有SKIPIF1<0SKIPIF1<0由不等式的傳遞性可得：SKIPIF1<0，即SKIPIF1<02、證明一元不等式主要的方法有兩個(gè)：第一個(gè)方法是將含SKIPIF1<0的項(xiàng)或所有項(xiàng)均挪至不等號(hào)的一側(cè)，將一側(cè)的解析式構(gòu)造為函數(shù)，通過(guò)分析函數(shù)的單調(diào)性得到最值，從而進(jìn)行證明，其優(yōu)點(diǎn)在于目的明確，構(gòu)造方法簡(jiǎn)單，但對(duì)于移項(xiàng)后較復(fù)雜的解析式則很難分析出單調(diào)性第二個(gè)方法是利用不等式性質(zhì)對(duì)所證不等式進(jìn)行等價(jià)變形，轉(zhuǎn)化成為SKIPIF1<0的形式，若能證明SKIPIF1<0，即可得：SKIPIF1<0，本方法的優(yōu)點(diǎn)在于對(duì)SKIPIF1<0的項(xiàng)進(jìn)行分割變形，可將較復(fù)雜的解析式拆成兩個(gè)簡(jiǎn)單的解析式。但缺點(diǎn)是局限性較強(qiáng)，如果SKIPIF1<0與SKIPIF1<0不滿足SKIPIF1<0，則無(wú)法證明SKIPIF1<0。所以用此類方法解題的情況不多，但是在第一個(gè)方法失效的時(shí)候可以考慮嘗試此法。3、在構(gòu)造函數(shù)時(shí)把握一個(gè)原則：以能夠分析導(dǎo)函數(shù)的符號(hào)為準(zhǔn)則。4、若在證明SKIPIF1<0中，解析式SKIPIF1<0可分解為幾個(gè)因式的乘積，則可對(duì)每個(gè)因式的符號(hào)進(jìn)行討論，進(jìn)而簡(jiǎn)化所構(gòu)造函數(shù)的復(fù)雜度。5、合理的利用換元簡(jiǎn)化所分析的解析式。6、判斷解析式符號(hào)的方法：（1）對(duì)解析式進(jìn)行因式分解，將復(fù)雜的式子拆分為一個(gè)個(gè)簡(jiǎn)單的式子，判斷出每個(gè)式子的符號(hào)即可得到解析式的符號(hào)（2）將解析式視為一個(gè)函數(shù)，利用其零點(diǎn)（可猜出）與單調(diào)性（利用導(dǎo)數(shù)）可判斷其符號(hào)（3）將解析式中的項(xiàng)合理分組，達(dá)到分成若干正項(xiàng)的和或者若干負(fù)項(xiàng)的和的結(jié)果，進(jìn)而判斷出解析式符號(hào)二、典型例題：例1：求證：SKIPIF1<0思路：移項(xiàng)構(gòu)造函數(shù)求解即可證明：所證不等式等價(jià)于：SKIPIF1<0令SKIPIF1<0則只需證明：SKIPIF1<0SKIPIF1<0令SKIPIF1<0解得：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0↗↘SKIPIF1<0SKIPIF1<0即所證不等式成立小煉有話說(shuō)：（1）此題的解法為證明一元不等式的基本方法，即將含SKIPIF1<0的項(xiàng)移至不等號(hào)的一側(cè)，構(gòu)造函數(shù)解決。（2）一些常見(jiàn)不等關(guān)系可記下來(lái)以備使用：①SKIPIF1<0②SKIPIF1<0③SKIPIF1<0例2：設(shè)函數(shù)SKIPIF1<0，證明：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0思路：本題依然考慮構(gòu)造函數(shù)解決不等式，但如果僅僅是移項(xiàng)，則所證不等式為SKIPIF1<0，令SKIPIF1<0，其導(dǎo)函數(shù)比較復(fù)雜（也可解決此題），所以考慮先對(duì)不等式進(jìn)行等價(jià)變形，轉(zhuǎn)變?yōu)樾问捷^為簡(jiǎn)單的不等式，再構(gòu)造函數(shù)進(jìn)行證明證明：SKIPIF1<0SKIPIF1<0SKIPIF1<0，所以所證不等式等價(jià)于SKIPIF1<0設(shè)SKIPIF1<0SKIPIF1<0只需證SKIPIF1<0即可SKIPIF1<0令SKIPIF1<0SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增SKIPIF1<0SKIPIF1<0故不等式得證小煉有話說(shuō)：本題在證明時(shí)采取先化簡(jiǎn)再證明的策略，這也是我們解決數(shù)學(xué)問(wèn)題常用的方法之一，先把問(wèn)題簡(jiǎn)單化再進(jìn)行處理。在利用導(dǎo)數(shù)證明不等式的問(wèn)題中，所謂的“簡(jiǎn)化”的標(biāo)準(zhǔn)就是構(gòu)造的函數(shù)是否易于分析單調(diào)性。例3：已知函數(shù)SKIPIF1<0，證明：SKIPIF1<0思路：若化簡(jiǎn)不等式左邊，則所證不等式等價(jià)于SKIPIF1<0，若將左邊構(gòu)造為函數(shù)，則函數(shù)的單調(diào)性難于分析，此法不可取?？紤]原不等式為乘積式，且與0進(jìn)行比較，所以考慮也可分別判斷各因式符號(hào)，只需讓SKIPIF1<0與SKIPIF1<0同號(hào)即可。而SKIPIF1<0的正負(fù)一眼便可得出，SKIPIF1<0的符號(hào)也不難分析，故采取分別判斷符號(hào)的方法解決。解：SKIPIF1<0SKIPIF1<0SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增SKIPIF1<0SKIPIF1<0為增函數(shù)SKIPIF1<0SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0SKIPIF1<0綜上所述，SKIPIF1<0成立小煉有話說(shuō)：與0比較大小也可看做是判斷一側(cè)式子的符號(hào)，當(dāng)不等式的一側(cè)可化為幾個(gè)因式的乘積時(shí)，可分別判斷每一個(gè)因式的符號(hào)（判斷相對(duì)簡(jiǎn)單），再?zèng)Q定乘積的符號(hào)。例4：已知SKIPIF1<0，其中常數(shù)SKIPIF1<0（1）當(dāng)SKIPIF1<0時(shí)，求函數(shù)SKIPIF1<0的極值（2）求證：SKIPIF1<0解：（1）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0在SKIPIF1<0單調(diào)遞增SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增SKIPIF1<0SKIPIF1<0的極小值為SKIPIF1<0，無(wú)極大值（2）思路：本題如果直接構(gòu)將左側(cè)構(gòu)造函數(shù)，則導(dǎo)數(shù)過(guò)于復(fù)雜，不易進(jìn)行分析，所以考慮將所證不等式進(jìn)行變形成“SKIPIF1<0”的形式。由第（1）問(wèn)可得：SKIPIF1<0，即SKIPIF1<0，則所證不等式兩邊同時(shí)除以SKIPIF1<0，即證：SKIPIF1<0，而SKIPIF1<0，所以只需構(gòu)造函數(shù)證明SKIPIF1<0即可解：由（1）得SKIPIF1<0所證不等式：SKIPIF1<0SKIPIF1<0設(shè)SKIPIF1<0SKIPIF1<0令SKIPIF1<0可解得：SKIPIF1<0SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減SKIPIF1<0SKIPIF1<0即SKIPIF1<0SKIPIF1<0例5：已知SKIPIF1<0（1）當(dāng)SKIPIF1<0時(shí)，求SKIPIF1<0在SKIPIF1<0的最值（2）求證：SKIPIF1<0，SKIPIF1<0解：（1）SKIPIF1<0SKIPIF1<0的單調(diào)區(qū)間為SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0↘↗SKIPIF1<0①SKIPIF1<0SKIPIF1<0SKIPIF1<0②SKIPIF1<0時(shí)，SKIPIF1<0（2）思路：所證不等式SKIPIF1<0，若都移到左邊構(gòu)造函數(shù)，則函數(shù)SKIPIF1<0很難分析單調(diào)性，進(jìn)而無(wú)法求出最值。本題考慮在兩邊分別求出最值，再比較大小即可解：所證不等式等價(jià)于SKIPIF1<0SKIPIF1<0設(shè)SKIPIF1<0SKIPIF1<0令SKIPIF1<0SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增SKIPIF1<0設(shè)SKIPIF1<0SKIPIF1<0SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0所證不等式成立例6：設(shè)SKIPIF1<0為常數(shù)），曲線SKIPIF1<0與直線SKIPIF1<0在(0，0)點(diǎn)相切.(1)求SKIPIF1<0的值.(2)證明：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.解：（1）SKIPIF1<0過(guò)SKIPIF1<0點(diǎn)SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0（2）思路：所證不等式等價(jià)于SKIPIF1<0，若將SKIPIF1<0的表達(dá)式挪至不等號(hào)一側(cè)，則所構(gòu)造的函數(shù)SKIPIF1<0中SKIPIF1<0，SKIPIF1<0求導(dǎo)后結(jié)構(gòu)比較復(fù)雜。觀察到對(duì)數(shù)與根式均含有SKIPIF1<0，進(jìn)而考慮換元SKIPIF1<0化簡(jiǎn)不等式。另一方面，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，而SKIPIF1<0是所證SKIPIF1<0的臨界值，進(jìn)而會(huì)對(duì)導(dǎo)數(shù)值的符號(hào)有所影響。解：所證不等式等價(jià)于：SKIPIF1<0令SKIPIF1<0則不等式轉(zhuǎn)化為：SKIPIF1<0SKIPIF1<0（若不去分母，導(dǎo)函數(shù)比較復(fù)雜，不易分析）令SKIPIF1<0SKIPIF1<0只需證SKIPIF1<0即可觀察SKIPIF1<0SKIPIF1<0SKIPIF1<0進(jìn)而考慮SKIPIF1<0的單調(diào)性（盡管SKIPIF1<0復(fù)雜，但有零點(diǎn)在，就能夠幫助繼續(xù)分析，堅(jiān)持往下進(jìn)行）SKIPIF1<0SKIPIF1<0單調(diào)遞增，SKIPIF1<0SKIPIF1<0單調(diào)遞減SKIPIF1<0（SKIPIF1<0是SKIPIF1<0的零點(diǎn)，從而引發(fā)連鎖反應(yīng)）SKIPIF1<0單調(diào)遞減SKIPIF1<0SKIPIF1<0即所證不等式成立SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0小煉有話說(shuō)：本題有以下兩個(gè)亮點(diǎn)（1）利用換元簡(jiǎn)化所證不等式（2）零點(diǎn)的關(guān)鍵作用：對(duì)于化簡(jiǎn)后的函數(shù)SKIPIF1<0而言，形式依然比較復(fù)雜，其導(dǎo)函數(shù)也很難直接因式分解判斷符號(hào)，但是由于尋找到SKIPIF1<0這個(gè)零點(diǎn)，從而對(duì)導(dǎo)函數(shù)的符號(hào)判斷指引了方向，又因?yàn)榘l(fā)現(xiàn)SKIPIF1<0也是導(dǎo)函數(shù)的零點(diǎn)，于是才決定在對(duì)導(dǎo)函數(shù)求一次導(dǎo)，在二次導(dǎo)函數(shù)中判斷了符號(hào)，進(jìn)而引發(fā)連鎖反應(yīng)，最終證明不等式?？梢哉f(shuō)，本題能堅(jiān)持對(duì)SKIPIF1<0進(jìn)行分析的一個(gè)重要原因就是SKIPIF1<0這個(gè)零點(diǎn)。例7：（，福建，20）已知函數(shù)SKIPIF1<0（1）求證：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0（2）求證：當(dāng)SKIPIF1<0時(shí)，存在SKIPIF1<0，使得對(duì)任意的SKIPIF1<0，恒有SKIPIF1<0解：（1）思路：所證不等式為：SKIPIF1<0，只需將含SKIPIF1<0的項(xiàng)移植不等號(hào)一側(cè)，構(gòu)造函數(shù)即可證明證明：所證不等式等價(jià)于：SKIPIF1<0，設(shè)SKIPIF1<0SKIPIF1<0SKIPIF1<0在SKIPIF1<0單調(diào)遞減SKIPIF1<0時(shí)，SKIPIF1<0即SKIPIF1<0得證（2）思路：本題的目標(biāo)是要找到與SKIPIF1<0相關(guān)的SKIPIF1<0，因?yàn)镾KIPIF1<0函數(shù)形式較為簡(jiǎn)單，所以可以考慮移至不等號(hào)一側(cè)：SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以只需SKIPIF1<0在SKIPIF1<0單增即可。可對(duì)SKIPIF1<0進(jìn)行SKIPIF1<0和SKIPIF1<0分類討論。證明：SKIPIF1<0設(shè)SKIPIF1<0則SKIPIF1<0且SKIPIF1<0令SKIPIF1<0，即SKIPIF1<0①當(dāng)SKIPIF1<0時(shí)，解得SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0恒成立SKIPIF1<0在SKIPIF1<0單調(diào)遞增SKIPIF1<0SKIPIF1<0可取任意正數(shù)②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0可取任意正數(shù)③當(dāng)SKIPIF1<0時(shí)，解得SKIPIF1<0，而SKIPIF1<0SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減SKIPIF1<0，均有SKIPIF1<0，只需取SKIPIF1<0即可綜上所述：存在SKIPIF1<0，使得對(duì)任意的SKIPIF1<0，恒有SKIPIF1<0例8：已知函數(shù)SKIPIF1<0（SKIPIF1<0為常數(shù)，SKIPIF1<0，是自然對(duì)數(shù)的底數(shù)），曲線SKIPIF1<0在SKIPIF1<0處的切線與SKIPIF1<0軸平行（1）求SKIPIF1<0的值（2）設(shè)SKIPIF1<0，其中SKIPIF1<0為SKIPIF1<0的導(dǎo)函數(shù)。證明：對(duì)SKIPIF1<0解：（1）SKIPIF1<0SKIPIF1<0SKIPIF1<0處的切線與SKIPIF1<0軸平行SKIPIF1<0SKIPIF1<0:（2）所證不等式等價(jià)于：SKIPIF1<0SKIPIF1<0設(shè)SKIPIF1<0SKIPIF1<0令SKIPIF1<0SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減SKIPIF1<0，即SKIPIF1<0若要證SKIPIF1<0，只需證SKIPIF1<0設(shè)SKIPIF1<0SKIPIF1<0，令SKIPIF1<0解得：SKIPIF1<0SKIPIF1<0在SKIPIF1<0單調(diào)遞增SKIPIF1<0SKIPIF1<0SKIPIF1<0，即原不等式得證例9：已知函數(shù)SKIPIF1<0，函數(shù)SKIPIF1<0的導(dǎo)函數(shù)SKIPIF1<0，且SKIPIF1<0，其中SKIPIF1<0為自然對(duì)數(shù)的底數(shù)．（1）求SKIPIF1<0的極值；（2）當(dāng)SKIPIF1<0時(shí)，對(duì)于SKIPIF1<0，求證：SKIPIF1<0．解：（1）函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上為增函數(shù)，SKIPIF1<0沒(méi)有極值；當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0SKIPIF1<0在SKIPIF1<0單調(diào)增，在SKIPIF1<0單調(diào)遞減SKIPIF1<0有極大值SKIPIF1<0，無(wú)極小值（2）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，令SKIPIF1<0，即SKIPIF1<0SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上為增函數(shù)SKIPIF1<0SKIPIF1<0SKIPIF1<0在SKIPIF1<0上為增函數(shù)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0，由SKIPIF1<0可知SKIPIF1<0SKIPIF1<0SKIPIF1<0即SKIPIF1<0例10：設(shè)函數(shù)SKIPIF1<0.（1）證明：SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；（2）證明：SKIPIF1<0.解：（1）SKIPIF1<0只需證SKIPIF1<0即可令SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0在SKIPIF1<0單調(diào)遞增SKIPIF1<0即SKIPIF1<0SKIPIF1<0函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增（2）思路：對(duì)所證不等式SKIPIF1<0，若直接將左側(cè)構(gòu)造函數(shù)，則無(wú)法求出單調(diào)區(qū)間和最值。（導(dǎo)函數(shù)中含有SKIPIF1<0無(wú)法進(jìn)一步運(yùn)算），所以考慮將左側(cè)的一部分挪至不等號(hào)另一側(cè)，構(gòu)造兩個(gè)函數(shù)進(jìn)行比較。SKIPIF1<0(右邊SKIPIF1<0,