以看做是關(guān)于r的一次函數(shù),6,7,8時(shí)可以看做是關(guān)于r的二次函數(shù),這樣最值一定在r的最值時(shí)取到,只要r的最值即可1r的一次函數(shù)的時(shí)候,r0或者兩數(shù)相等的時(shí)候取到(此證明見(jiàn)<theuvwmethod>) 2、當(dāng)三元不等式轉(zhuǎn)化成關(guān)于r的二次函數(shù)的時(shí)候,只需考慮此二次函數(shù)的開(kāi)口和對(duì)稱r在何處取到最值66pqr(a5bab5)(5,1)(a4b2a2b4)(4,a3b3(3,3)q3a4bc(4,1,1)(a3b2ca3bc2)a2b2c2(2,2,2)r1abc0求證

a2b2abbc

2(ab)(bc)(ca)p2

pabc1,qabbcca,rabc 2pq3 pq

pr11qrf(r11qp2rp3q3pq2此為一次函數(shù)f(r)()rq3q2此為關(guān)于r的一次函數(shù),由pqr結(jié)論可知最值在一數(shù)為0或兩數(shù)相等時(shí)取到情況1一數(shù)為0a0,f(0)bc3(bc)2g(bc注意到bcbc)21g(bc)是開(kāi)口向下的二次 最小值在邊界取到g(0)0g1)

0均大于等于 情況2:兩數(shù)相abt,c12t;f(r)[11(t22t4t2)1]t2(12t)t22t4t23(t22t4t266x5104x460x316x22xh(x)(0x1)2x(33x452x330x28x1)2l(x)132x3156x260x84(33x339x215x2)4[(6x33x2)(27x336x215x12x2(2x1)4(3x1)2(3x2)12x2(11)4(3x1)2(1)2sol(x)l1)0.06250f(r)h(x)2xl(x)0均大于等于0綜上,原不等式得2:a,b,c0,abc3,max{(a2abb2)(b2bcc2)(c2caa2)}做代換pabc3qabbccar第一題:5r227q81)r9q2對(duì)于每一個(gè)能取到的q值,都可以把上式記做關(guān)于r的二次函注意到3qp29q3so對(duì)稱軸27q81由于r0so此開(kāi)口向下的二次函數(shù)的最大值在0處取到不妨設(shè)c0帶入原式:得到:(a2abb2)(b2bcc2)(c2caa2)(a2abb2 22

2

a22abb2 4(ab再由均值:(aabb)ab=(aabb)(ab)(ab) ) 9第二題:3r27)r9q2對(duì)于每一個(gè)能取到的q值都可以把上式看做關(guān)于的二次函數(shù),的二次函數(shù)3p3說(shuō)明函數(shù)的定義域?yàn)?r3對(duì)于取定

,由均值q33r2r最大值只能在r0或r

q處取33當(dāng)r

時(shí),根據(jù)均值的取等條件,只能是q3abc1此時(shí)原式的值為

當(dāng)r0時(shí)不妨設(shè)c0,帶入原式得到(a2abb2)(b2bcc2)(c2caa2)(a2abb2 2

ab

3

a22abb2

39

(aabb)ab (ab )(2ab) ) 4 44 注意到218727,so最大值2187,當(dāng)且僅當(dāng)ab3c0取 r=√(q3/27)時(shí)，q=3，a=b