基本不等式The basic Inequalities均值不等式AM-GM InequalityAM-GM 不等式及其应用AM-GM Inequality and ApplicationsAM-GM 不等式the:AM-GM 不等 对所有正实数 a 1,a 2,⋯,a n ,下列不等式成立a 1+a 2+⋯+a n n≥√a 1a 2⋯a n n 当且仅当 a 1=a 2=⋯=a n 时, 等号成立.For all positive real numbers a 1,a 2,…,a n , the following inequality holdsa 1+a 2+⋯+a n n≥√a 1a 2⋯a n n Equality occurs if and only if a 1=a 2=⋯=a n .Proof. 当 n =2 时,不等式显然成立. 如果不等式对 n 个正实数成立,那么它对 2n 个正实数也成立. 这是由于a 1+a 2+⋯+a 2n ≥n √a 1a 2⋯a n n +n √a n+1a n+2⋯a 2n n ≥2n √a 1a 2⋯a 2n 2n因此不等式对 n 是 2 的指数幂形式个正实数是成立的. 假设不等式对 n 成立,我们设a n =s n −1,s =a 1+a 2+⋯+a n−1 依据归纳假设,我们有s +s n −1≥n √a 1a 2⋯a n−1⋅s n −1n ⇒s ≥(n −1)√a 1a 2⋯a n−1n−1 因此,如果不等式对 n 个正实数成立,那么它对 n −1 个正实数成立,由归纳法 (Cauchy 归纳) 可知,不等式对每一个自然数 n 都是成立的. 当且仅当 a 1= a 2=⋯=a n 时,等号成立. ◻Proof. The inequality is clearly true for n =2. If it is true for n numbers, it will be true for 2n numbers becausea 1+a 2+⋯+a 2n ≥n √a 1a 2…a n n +n √a n+1a n+2⋯a 2n n ≥2n √a 1a 2⋯a n 2nThus the inequality is true for every number n that is an exponent of 2 . Suppose that the inequality is true for n numbers. We then choosea n =s n −1;s =a 1+a 2+⋯+a n−1According to the inductive hypothesis, we gets+sn−1≥n√a1a2…a n−1⋅sn−1n⇒s≥(n−1)√a1a2…a n−1n−1Therefore if the inequality is true for n numbers, it will be true for n−1 numbers. By induction (Cauchy induction), the inequality is true for every natural number n. Equality occurs if and only if a1=a2=⋯=a n◻AM-GM 不等式作为一个著名的、应用广泛的定理,在证明不等式方面也是不可缺少的.下面通过一些著名的不等式来研究它的强大的应用.As a matter of fact, the AM-GM inequality is the most famous and wide-appbed theorem. It is also indispensable in proving inequalities. Consider its strong applications through the following famous inequalities.Nesbitt不等式pro:Nesbitt不等式 (1) 证明对所有非负实数a,b,c,ab+c +bc+a+c a+b ≥32.(2) 证明对所有非负实数a,b,c,d,ab+c +bc+d+cd+a+da+b≥2.Proof. (1) 考虑下列表达式S=ab+c+bc+a+ca+bM=bb+c+cc+a+aa+bN=cb+c+ac+a+ba+b我们有M+N=3, 根据 AM-GM 不等式,我们得到M+S=a+bb+c+b+cc+a+c+aa+b≥3N+S=a+cb+c+a+bc+a+b+ca+b≥3所以M+N+2S≥6, 即有2S≥3⇒S≥32.(2) 考虑下列表达式S=ab+c+bc+d+cd+a+da+bM=bb+c+cc+d+dd+a+aa+bN=cb+c+dc+d+ad+a+ba+b则M+N=4. 根据 AM-GM 不等式,我们有M+S=a+bb+c+b+cc+d+c+dd+a+d+aa+b≥4N+S=a+cb+c+b+dc+d+a+cd+a+b+da+b=a+cb+c+a+cd+a+b+dc+d+b+da+b≥4(a+c)a+b+c+d+4(b+d)a+b+c+d=4因此, M+N+2S≥8, 即2S≥4⇒S≥2.当且仅当a=b=c=d或a=c,b= d=0或a=c=0,b=d. ◻其他证法Proof 1 . Using the well-known inequality(x+y+z)(1x+1y+1z)≥9 for x,y,z>0We get (ab+c +1)+(bb+c+1)+(ca+b+1)=12[(a+b)+(b+c)+(c+a)](1a+b+1 b+c +cc+a)≥92Therefore, ab+c +bc+a+ca+b≥92−3=32.Proof 2. Setting x=b+c,y=c+a,z=a+b, thena=y+z−x2,b=z+x−y2,c=x+y−z2(1) can be written asy+z−x2x +z+x−y2y+x+y−z2z≥32⇔(xy+yx)+(yz+zy)+(zx+xz)≥6This inequality is clearly true by the well-known inequality pq +qp≥2 for p,q>0Proof 3. (1) is equivalent to2[a(a+b)(a+c)+b(b+a)(b+c)+c(c+a)(c+b)]≥3(a+b)(b+c)(c+a)⇔2(a3+b3+c3)≥ab(a+b)+bc(b+c)+ca(a+b)The inequality follows by the well-known inequalityx3+y3≥xy(x+y)(⇔(x+y)(x−y)2≥0) for x,y≥0 Proof 4 . Using the AM-GM inequality, we obtaina2 b+c +b+c4≥2√a2b+c⋅b+c4=aSimilarly, we get b2c+a +c+a4≥b, and c2a+b+a+b4≥c.Adding three above inequalities, we havea2 b+c +b2b+c+c2a+b+a+b+c2≥a+b+c⇔(a2b+c +a)+(b2b+c+b)+(c2a+b+c)≥32(a+b+c)⇔a(a+b+c)b+c+b(a+b+c)b+c+c(a+b+c)a+b≥32(a+b+c)⇔ab+c+bc+a+ca+b≥32.Proof 5. SettingA=ab+c+bc+a+ca+b,B=bb+c+cc+a+aa+b,C=cb+c+ac+a+ba+bthenB+C=3,A+B=a+bb+c+b+cc+a+c+aa+b,A+C=a+cb+c+b+ac+a+c+ba+b.By the AM-GM inequality, we haveA+B≥3√a+bb+c⋅b+cc+a⋅c+aa+b3=3,A +C ≥3√a +cb +c ⋅b +a c +a ⋅c +b a +b3=3 Thus, 2A +B +C ≥6 or A ≥32. Proof 6 . By the AM-GM inequality, we have√a 3+√b 3+√b 3≥3√√a 3b 63=3b √a√a 3+√c 3+√c 3≥3√√a 3c 63=3c √a. Adding two inequalities, we get2(√a 3+√c 3+√c 3)≥3√a (b +c )Thus,a b +c ≥32⋅√a 3√a 3+√b 3+√c 3.Similarly, we haveb c +a ≥32⋅√b 3√a 3+√b 3+√c 3,c a +b ≥32⋅√c 3√a 3+√b 3+√c 3. Adding three above inequalities, we obtain Adding three above inequalities, we obtaina b +c +b c +a +c a +b ≥32⋅√a 3+√b 3+√c 3√a 3+√b 3+√c 3=32Proof 7 . Using the well-known inequality(x +y )2≥4xy, for x,y ≥0We have[2a +(b +c )]2≥4⋅2a (b +c ) or 4a 2+4a (b +c )+(b +c )2≥8a (b +c ) Hence,4a (a +b +c )≥(b +c )[8a −(b +c )]⇒a b +c ≥14⋅8a −b −c a +b +c.Similarly, we getb c+a ≥14⋅8b−c−aa+b+c,c a+b ≥14⋅8c−a−ba+b+c.Adding three inequalities. we havea b+c +bc+a+ca+b≥14⋅6(a+b+c)a+b+c=32Proof 8. (By Cao Minh Quang) Using the AM-GM inequality, we have2(a+b+c)3=2a+(b+c)+(b+c)3≥√2a(b+c)23.Thus,a b+c ≥3√32⋅a√a√(a+b+c)3.Similarly, we getb c+a ≥3√32⋅b√b√(a+b+c)3,c a+b ≥3√32c√c√(a+b+c)3.Adding three inequalities, we havea b+c +bc+a+ca+b≥3√32⋅a√a+b√b+c√c√(a+b+c)3We have to show that3√3 2⋅a√a+b√b+c√c√(a+b+c)3≥32⇔3(a√a+b√b+c√c)2≥(a+b+c)3 (2) We set x=√a,y=√b,z=√c.(2) can be written as3(x3+y3+z3)2≥(x2+y2+z2)3By the Cauchy-Schwarz inequality, we have(x2+y2+z2)2≤(x3+y3+z3)(x+y+z) (3)By the Chebyshev’s inequality, we have(x2+y2+z2)(x+y+z)≤3(x3+y3+z3) (4) Multiplying (3) and (4) yields, we obtain3(x3+y3+z3)2≥(x2+y2+z2)3 Proof 9 . (1) is equivalent to(ab+c −12)+(bc+a−12)+(ca+b−12)≥0⇔a−b+a−cb+c+b−c+b−ab+c+c−a+c−ba+b≥0⇔(a−b)(1b+c−1c+a)+(b−c)(1a+c−1c+b)+(a−c)(1b+c−1a+b)≥0⇔(a−b)2(b+c)(a+c)+(b−c)2(a+c)(a+b)+(c−a)2(b+c)(a+b)≥0.The last inequality is clearly true.Proof 10. By the Cauchy - Schwarz inequality, we havea b+c +bc+a+ca+b≥(a+b+c)2a(b+c)+b(c+a)+c(c+a) =(a+b+c)22(ab+bc+ca)≥3(ab+bc+ca)2(ab+bc+ca)=32Proof 11 . Without loss of generality, we can assume that a≥b≥c, then 1b+c ≥1 c+a ≥1a+b.By the Chebyshev’s inequality and the AM-GM inequality, we havea b+c +bc+a+ca+b≥13(a+b+c)(1b+c+1c+a+1a+b)=16[(a+b)+(b+c)+(c+a)](1a+b+1b+c+1c+b)≥96=32Proof 12. Setting x=ab+c ,y=bc+a,z=ca+b, and A=x+y+z3. We eed to prove that A≥12. We havex 1+x +y1+y+z1+z=aa+b+c+ba+b+c+ca+b+c=1Thus, 1=2xyz+xy+yz+zx By the AM-GM inequality, we get1=2xyz +xy +yz +zx ≤2A 3+3A 3⇒(2A −1)(A +1)2≥0Since A >0, hence A ≥12.Proof 13. Using the same substitution as the 12tℎ proof, and setting f (t )= t 1+t . It is easy to show that f (t ) is increase and concave on (0,+∞). By the Jensen’s inequality, we havef (12)=13=13[f (x )+f (y )+f (z )]≤f (x +y +z 3)⇒12≤x +y +z 3 or 32≤a b +c +b c +a +c a +b Proof 14. (By Cao Minh Quang) Firstly, we state and prove a lemma.Lemma. If x i ,y i ,(i =1,2,3) are positive real numbers satisfy that x 1≥ x 2≥x 3,y 1≥y 2≥y 3, thenx 1y 1+x 2y 2+x 3y 3≥x 1y i 1+x 2y i 2+x 3y i 3 (5)where (i 1,i 2,i 3) is a permutation of (1,2,3)Proof (5). We set z 1=y i 1,z 2=y i 2,z 3=y i 0. (5) becomesx 1y 1+x 2y 2+x 3y 3≥x 1z 1+x 2z 2+x 3z 3⇒x 1(y 1−z 1)+x 2(y 2−z 2)+x 3(y 3−z 3)≥It is easy to see that y 1≥z 1,y 1+y 2≥z 1+z 2 and y 1+y 2+y 3=z 1+z 2+z 3. Therefore,x 1(y 1−z 1)+x 2(y 2−z 2)+x 3(y 3−z 3)≥x 2(y 1−z 1)+x 2(y 2−z 2)++x 3(y 3−z 3)=x 2[(y 1+y 2)−(z 1+z 2)]+x 3(y 3−z 3)≥x 3[(y 1+y 2)−−(z 1+z 2)]+x 3(y 3−z 3)=x 3[(y 1+y 2+y 3)−(z 1+z 2+z 3)]=0Let us now prove (1). Without loss of generality, we can assume that a ≥b ≥c , then 1b+c ≥1c+a ≥1a+b. Using (5), we get a b+c+b c+a +c a+b ≥b b+c +c c+a +a a+b and a b +c +b c +a +c a +b ≥c b +c +a c +a +b a +bAdding two inequalities, we obtain 2(a b+c +b c+a +c a+b )≥b+c b+c +c+a c+a +a+b a+b =3 ora b +c +b c +a +c a +b ≥32Proof 15 . Without loss of generality, we can assume that a≥b≥c. We set x=a c ,y=bc, then x≥y≥1(1) becomesx y+1+yx+1+1x+y≥32 (6)Using the AM-GM inequality, we havex+1 y+1+y+1x+1≥2⇒xy+1+yx+1≥2−1x+1−1y+1It suffices to prove that2−1x+1−1y+1≥32−1x+y⇔12−1y+1≥1x+1−1x+y⇔y−12(y+1)≥y−1(x+1)(x+y)⇔(y−1)[(x+1)(x+y)−2(y+1)]2(x+y)(x+1)(y+1)≥0The last inequality i true since x≥y≥1To prove (6), beside the above proof, we also have an another proof. Proof 16. We set m=x+y,n=xy (6) becomesm2−2n+m m+n+1+1m≥32⇔2m3−m2−m+2≥n(7m−2)We note that 7m−2>0 and m2≥4n. It suffices to prove that4(2m3−m2−m+2)≥m2(7m−2)⇔(m−2)2(m+2)≥0 This inequality is clearly true.Proof 17. (By Cao Minh Quang)By setting x=ab ,y=bc,z=ca, then xyz=1.(1) can be written asx xz+1+yyx+1+zzy+1≥32⇔2(x2y+y2z+z2x)≥(x+y+z)+(xy+yz+zx) (7) Using the AM-GM inequality, we have(x2y+y2z+z2x)=13[(x2y+y2z+y2z)+(y2z+z2x+z2x)+(z2x+x2y+x2y)]≥√x2y5z23+√y2z5x23+√z2x5y23=x+y+zUsing the AM-GM inequality again, we have(x2y+y2z+z2x)=13[(x2y+z2x+z2x)+(y2z+x2y+x2y)+(z2x+y2z+y2z)]≥√x4z4y3+√x4y4z3+√xy4z43=xz+xy+yzAdding two inequalities, we get (7)Since (1) in the homogeneous, we can assume that a+b+c=1. We have to provea b+c +bc+a+ca+b≥32, where a+b+c=1 (1′)We have some proofs of (1′)Proof 18 . For 0<x<1, we note that4x−(1−x)(9x−1)=(3x−1)2Hence, 4x≥(1−x)(9x−1) or x1−x ≥9x−14Therefore,a b+c +bc+a+ca+b=a1−a+b1−b+c1−c≥≥9a−14+9b−14+9c−14=9(a+b+c)−34=32Proof 19. (By Cao Minh Quang) For 0<x<1, we note that(3x−1)2≥0⇔3x+1≥9x(1−x)Hence 11−x ≥9x3x+1or x1−x≥9x23x+1Therefore, by Cauchy-Schwarz Inequality, we havea b +c +b c +a +ca +b=a 1−a +b 1−b +c 1−c ≥9a 23a +1+9b 23b +1+9c 23c +1≥(3a +3b +3c )23(a +b +c )+3=32Proof 20 . (By Cao Minh Quang) For 0<x <1, by the AM-GM inequality, we get(2−2x )(1+x )(1+x )≤[(2−2x )+(1+x )+(1+x )3]3=6427Therefore, x1−x ≥2732x (1+x )2. Using the well-known inequalityx r +y r +z r 3≥(x +y +z 3)r, for x,y,z ≥0,r ≥1We geta b +c +b c +a +c a +b=a 1−a +b 1−b +c1−c ≥2732∑a a,b,c(1+a )2=2732[∑a 3a,b,c+2∑a 2a,b,c+∑a a,b,c]≥2732[3(13)3+6(13)2+1]=32Proof 21 . (By Cao Minh Quang) By the AM-GM inequality, wea b +c +9a (b +c )4≥2√a b +c ⋅9a (b +c )4=3aSimilarly, we have bc+a +9b (c+a )4≥3b,c a+b +9c (a+b )4≥3c.Adding three inequalities and noting that(a +b +c )=(a +b +c )2≥3(ab +bc +ca )We obtaina b +c +b c +a +ca +b≥3(a +b +c )−92(ab +bc +ca )=32(a +b +c )+[32(a +b +c )−92(ab +bc +ca )]≥32Proof 22. Setting f (t )=t1−t . It is easy to show that f (t ) is a convex function on (0,1). By the Jensen’s inequality, we obtain f (a )+f (b )+f (c )≥3f (a +b +c3) or a 1−a +b 1−b +c 1−c ≥32Proof 23 . Without loss of generality, we can assume that a ≥b ≥c. We get a ≥13,c ≤13, which follows that(a,b,c )≽(13,13,13). Since f (t )=t1−t is a convex function on (0,1), by the Karamata’s inequality, we obtainf (a )+f (b )+f (c )≥f (13)+f (13)+f (13)or a1−a +b1−b +c1−c ≥32. We are done.Proof 24. (By Cao Minh Quang) To prove (1’), we need to prove a lemma. Lemma. (Cao Minh Quang and Tran Tuan Anh)If a,b,c are positive real numbers such that a +b +c =1, thena +bcb +c +b +ca c +a +c +aba +b≥2 Proof. Using the well-known inequalityx 2+y 2+z 2≥xy +yz +zx,for x,y,z >0We geta +bcb +c +b +ca c +a +c +aba +b=(a +b )(a +c )b +c +(b +c )(b +a )c +a +(c +a )(c +b )a +b≥(a +b )+(b +c )+(c +a )=2We now prove (1’). Using the well -known inequalityxy x +y ≤14(x +y ), for x,y >0 We obtaina b +c +b c +a +ca +b≥2−(bc b +c +ca c +a +aba +b )≥≥2−14[(b +c )+(c +a )+(a +b )]=32Last, we use "Mixing Variables Theorem" to prove (1). Proof 25. By setting E (a,b,c )=ab+c +bc+a +ca+b , t =a+b 2, and v =a+b+c 3. It is easy tosee thatE (a,b,c )=a 2+b 2+c (a +b )ab +c 2+c (a +b )+c a +b ≥2t 2+2tc t 2+c 2+2ct +c2t =E (t,t,c )Thus, E (a,b,c )≥E (v,v,v )=32.加权AMGM 不等式pro:加权AMGM 不等式 假设 a 1,a 2,⋯,a n 是正实数, 如果 n 个非负实数 x 1,x 2,⋯,x n 的和为 1 , 则a 1x 1+a 2x 2+⋯+a n x n ≥a 1∗1a 2x 2⋯a n xnProof. 这个不等式的证明和经典 AM-GM 不等式的证明是类似的.在 n =2 的情形,我们必须详细地证明(因为不等式中出现了实数指数). 我们先来证明, 如果 x,y ≥0,x +y =1 以及 a,b >0, 则ax +by ≥a x b y证明这个不等式的最简单的方法是考虑 x,y 是有理数的情况,至于实数我们可以采用极限的方法来进行. 如果 x,y 是有理数,设 x =m m+n ,y =nm+n (m,n ∈N ), 根据 AM-GM 不等式, 我们有ma+nb≥(m+n)amm+n bnm+n⇒ax+by≥a x b y如果x,y是实数,则存在两个有理数序列{r n},{s n}(n≥0,n∈N),使得r n→x,s n→y,r n+s n=1于是ar n+bs n≥a r n b s n或者ar n+b(1−r n)≥a r n b1−r n取极限,令n→+∞, 我们有ax+by≥a x b y◻尽管 AM-GM 不等式非常简单,但它在数学竞赛中的不等式证明方面扮演着重要的角色.用下面的一些例子来帮助你熟悉这个重要的不等式.设a,b,c>0且a+b+c=3, 证明√a+√b+√c≥ab+bc+caProof.注意到恒等式2(ab+bc+ca)=(a+b+c)2−(a2+b2+c2)则该不等式等价于∑a2 cyc +2∑√acye≥9由 AM-GM 不等式,我们有∑a2 cyc +2∑√aeyc=∑(a2+√a+√a)cyc≥3∑acyc=9所以,不等式成立. ◻(IMO Shortlist 1998) 设x,y,z>0, 且xyz=1, 证明x3(1+y)(1+z)+y3(1+z)(1+x)+z3(1+x)(1+y)≥34Proof.利用 AM-GM 不等式, 我们有x3(1+y)(1+z)+1+y8+1+z8≥3x4因此∑x3(1+y)(1+z)cyc +14∑(1+x)cyc≥∑3x4cyc⇒∑x3(1+y)(1+z)cyc ≥14∑(2x−1)cyc≥34当x=y=z=1时,等号成立. ◻( APMO 1998) 设x,y,z>0, 证明(1+xy)(1+yz)(1+zx)≥2+2(x+y+z)√xyz3Proof.不等式整理之后等价于x y +yz+zx≥x+y+z√xyz3由 AM-GM 不等式, 我们有3(xy +yz+zx)=(2xy+yz)+(2yz+zx)+(2zx+xy)≥3x√xyz3+3y√xyz3+3z√xyz3即x y +yz+zx≥x+y+z√xyz3◻设a,b,c,d>0, 证明16(abc+bcd+cda+dab)≤(a+b+c+d)3 Proof.应用 AM-GM 不等式, 我们得到16(abc+bcd+cda+dab)=16ab(c+d)+16cd(a+b)≤4(a+b)2(c+d)+4(c+d)2(a+b)=4(a+b+c+d)(a+b)(c+d)≤(a+b+c+d)3当且仅当a=b=c=d时,成立等号. ◻( Pham Kim Hung) 设a,b,c是周长为 3 的三角形的三边长,证明1√a+b−c +1√b+c−a+1√c+a−b≥9ab+bc+caProof.设x=√b+c−a,y=√c+a−b,z=√a+b−c,我们有x2+y2+z2=3不等式变为1 x +1y+1z≥369+x2y2+y2z2+z2x2又设m=xy,n=yz,p=zx, 则上述不等式等价于(m+n+p)(m2+n2+p2+9)≥36√mnp由 AM-GM 不等式, 我们有m+n+p≥3√mop3m2+n2+p2+9=m2+n2+p2+1+⋯+1≥12√m2⋅n2⋅p2⋅1⋅⋯⋅112=12√mnp6两式相乘即得(m+n+p)(m2+n2+p2+9)≥36√mnp所以,原不等式成立. ◻(Phan Thanh Nam) 设a1,a2,⋯,a n是正实数,且满足a i∈[0, i],i∈{1,2,⋯,n}, 证明2n a1(a1+a2)⋯(a1+a2+⋯+a n)≥(n+1)a12a22⋯a n2Proof.根据 AM-GM 不等式,我们有a1+a2+⋯+a k=1⋅(a11)+2⋅(a22)+⋯+k⋅(a kk)≥k(k+1)2(a11)2k(ℎ+1)⋅(a22)4k(k+1)⋯(a kk)2kk(k+1)将上述不等式对于k∈{1,2,⋯,n},相乘,我们有∏(a1+a2+⋯+a k) nk=1≥∏(k(k+1)2∏(a ii)2ik(k+1)ki=1) nk=1=n!(n+1)!2n∏(a ii)c ini=1这里的指数c i由下式确定c i=2i[1i(i+1)+1(i+1)(i+2)+⋯+1n(n+1)]=2i(1i−1n+1)≤2这是由于a i≤i,i∈{1,2,⋯,n},(a ii )c i≥(a ii)2.因此a1(a1+a2)⋯(a1+a2+⋯+a n)≥n!(n+1)!2n∏(a ii)2ni=1=n+12na12a22⋯a n2当且仅当a i=i(i=1,2,⋯,n)时,成立等号. ◻( USA MO 1998) 设a,b,c是正实数,证明1a3+b3+abc +1b3+c3+abc+1c3+a3+abc≤1abcProof.注意到a3+b3≥ab(a+b), 所以abca3+b3+abc ≤abcab(a+b)+abc=ca+b+c类似地可得另外两个不等式,并将它们相加,我们有abca3+b3+abc +abcb3+c3+abc+abcc3+a3+abc≤1所以,原不等式成立. ◻注意 IMO Shortlist 1996 类似的问题. 设x,y,z是积为 1 的三个正数, 证明xyx5+xy+y5+yzy5+yz+z5+zxz5+zx+x5≤1如果x1,x2,⋯,x n>0, 且满足11+x1+11+x2+⋯+11+x n=1,证明x1x2⋯x n≥(n−1)n Proof.所给条件变形为1 1+x1+11+x2+⋯+11+x n−1=x n1+x n应用AM - GM 不等式,我们有x n 1+x n ≥n−1√(1+x1)(1+x2)⋯(1+x n−1) n−1类似地可得其他n−1个不等式,并将它们相乘, 即得所证不等式. ◻设x,y,z是正数, 且满足x5+y5+z5=3, 证明x4 y3+y4z3+z4x3≥3Proof.注意到(x5+y5+z5)2=x10+2x5y5+y10+2y5z5+z10+2z5x5=9从这个形式, 我们利用AM−GM不等式,可得10⋅x4y3+6x5y5+3x10=x4y3+⋯+x4y310+x5y5+⋯+x5y5⏟6+x10+x10+x10≥19x10019类似地, 可得10⋅y4z3+6y5z5+3y10≥19y1001910⋅z4x3+6z5x5+3z10≥19z10019将上述三个不等式相加,我们有10(x4y3+y4z3+z4x3)+3(x5+y5+z5)2≥19(x10019+y10019+z10019)于是,只需证明下列不等式x 10019+y10019+z10019≥x5+y5+z5而这是成立的. 事实上(x5+y5+z5)+19∑x10019cyc =3+19∑x10019cyc=∑(1+19x10019)cyc≥20∑x5cyc◻(Mathlinks Contest) 设a,b,c>0,abc=1, 证明√a +b a +1+√b +c b +1+√c +a c +1≥3 Proof. 由AM-GM 不等式,我们有LHS ≥3√√a +b a +1⋅√b +c b +1⋅√c +ac +13=3√(a +b )(b +c )(c +a )(a +1)(b +1)(c +1)6 (∗)于是,我们只需证明(a +b )(b +c )(c +a )≥(a +1)(b +1)(c +1) (∗)由于 abc =1, 所以式 (∗) 等价于ab (a +b )+bc (b +c )+ca (c +a )≥a +b +c +ab +bc +ca根据 AM-GM 不等式,我们有2LHS +∑a cycb =∑(a 2b +a 2b +a 2c +a 2c +bc )cyc≥5∑acyc2LHS +∑a cyc=∑(a 2b +a 2b +b 2a +b 2a +c )cyc≥5∑a cycb由此4LHS +∑a cgcb +∑a cycb ≥5∑a cyc+5∑a cycb ⇒4LHS ≥4∑a cyc+4∑a cycb =4RHS所以, 原不等式成立. 当且仅当 a =b =c =1, 等号成立. ◻ 设 a,b,c 是某三角形的三边长,证明(a +b −c )a (b +c −a )b (c +a −b )c ≤a a b b c cProof. 由加权 AM-GM 不等式,我们有(a +b −c a )aa+b+c⋅(b +c −a b )b a+b+c⋅(c +a −bc )c a+b+r≤a a +b +c ⋅a +b −c a +b a +b +c ⋅b +c −a b +c a +b +c ⋅c +a −bc=1整理即得(a +b −c )a (b +c −a )b (c +a −b )c ≤a a b b c c当且仅当 a =b =c 时,成立等号. ◻设 a,b,c ≥0,a +b +c =2, 证明a 2b 2+b 2c 2+c 2a 2≤1Proof. 由恒等式(ab +bc +ca )(a 2+b 2+c 2)=ab (a 2+b 2)+bc (b 2+c 2)+ca (c 2+a 2)+abc (a +b +c ) 我们有(ab +bc +ca )(a 2+b 2+c 2)≥ab (a 2+b 2)+bc (b 2+c 2)+ca (c 2+a 2)≥2(a 2b 2+b 2c 2+c 2b 2) (1) 利用 AM-GM 不等式,以及a 2+b 2+c 2+2(ab +bc +ca )=(a +b +c )2=4我们有2(ab +bc +ca )(a 2+b 2+c 2)≤(2(ab +bc +ca )+(a 2+b 2+c 2)2)2≤4即(ab +bc +ca )(a 2+b 2+c 2)≤2 (2)由式 (1),(2) 即得a 2b 2+b 2c 2+c 2a 2≤1当且仅当 a =b =1,c =0 或其循环排列时,等号成立. ◻ 设 a,b,c,d >0, 证明1a 2+ab +1b 2+bc +1c 2+cd +1d 2+da ≥4ac +bdProof. 注意到ac +bd a 2+ab =a 2+ab +ac +bd a 2+ab −1=a (a +c )+b (d +a )a (a +b )−1=a +c a +b +b (d +a )a (a +b )−1 根据 AM-GM 不等式,我们得到(ac +bd )∑1a 2+ab cyc=∑a +c a +b cyc+∑b (d +a )a (a +b )cyc−4≥∑a +ca +bcyc此外∑a+c a+bcyc =(a+c)(1a+b+1c+d)+(b+d)(1b+c+1d+a)≥4(a+c)a+b+c+d+4(b+d)a+b+c+d=4当且仅当a=b=c=d时,等号成立. ◻设a,b,c,d,e≥0, 且a+b+c+d+e=5, 证明abc+bcd+cde+dea+eab≤5 Proof.不失一般性,我们设e=min(a,b,c,d,e).根据 AM-GM 不等式,我们有abc+bcd+cde+dea+eab=e(a+c)(b+d)+bc(a+d−e)≤e(a+b+c+d2)2+(b+c+a+d−e3)3=e(5−e)24+(5−2e)327因此,只需证明e(5−e)24+(5−2e)327≤5这是成立的, 可由(e−1)2(e+8)≥0推出. ◻(Pham Kim Hung) 设a,b,c,d>0, 证明(1a+1b+1c+1d)2≥1a2+4a2+b2+9a2+b2+c2+16a2+b2+c2+d2Proof.我们来证明1 b2+1c2+1d2+∑2absym≥4a2+b2+9a2+b2+c2+16a2+b2+c2+d2凷AM−GM不等式,我们有2 ab ≥4a2+b22 ac +2bc≥8ac+bc≥8a2+b2+c21 b2+1c2≥4b2+c2≥1b2+c2+a22 ad +2bd+2cd≥18ad+bd+cd≥16a2+b2+c2+d2将上述不等式相加,即得所要证明的不等式. ◻注意(1) 使用类似的方法,我们可以证明 5 个变量的类似的不等式. 为此,我们有a2+b2+c2+d2+e2=(a2+e24)+(b2+e24)+(c2+e24)+(d2+e24)≥ae+be+ce+de(2) 本题不等式加强为(1a+1b+1c+1d)2≥1a2+4a2+b2+12a2+b2+c2+18a2+b2+c2+d2(3) 猜想:设a1,a2,⋯,a n>0, 证明或否定(1a1+1a2+⋯+1a n)2≥1a12+4a12+a22+⋯+n2a12+a22+⋯+a n2(IMO 2006) 求不等式|ab(a2−b2)+bc(b2−c2)+ca(c2−a2)|≤M(a2+b2+c2)2对所有实数a,b,c都成立的最小的M值.记x=a−b,y=b−c,z=c−a,s=a+b+c, 则不等式可以写成如下形式9| sxyz∣≤M(s2+x2+y2+z2)2其中, s,x,y,z是任意实数,且满足x+y+z=0.事实上, s是一个独立变量. 首先我们来考察xyzz和x2+y2+z2之间的关系. 由于x+y+z=0, 很显然,x,y,z中有两个变量的符号相同,不妨设为x,y, 因此假定x,y≥0(x,y≤0类似可证). 由 AM-GM 不等式, 我们有|sxyz|=|sxy(x+y)|≤|s|⋅(x+y)34等号当x=y时成立. 设t=x+y, 再次应用 AM-GM 不等式,我们有2s2t6=2s2⋅t2⋅t2⋅t2≤(2s2+3t2)444于是4√2|s|t3≤(s2+32t2)2≤(s2+x2+y2+z2)2结合式(1),(2), 我们得到|sxyz |≤116√2s 2+x 2+y 2+z 2)2 这就意味着 M ≥9√232. 为证明 M =9√232 是最好的常数,我们必须求出 (s,x,y,z ), 即(a,b,c ), 经简单的计算,得到等号成立的值为(a,b,c )=(1−3√21,1+3√2) 使用 AM-GM 不等式的重要的原则是选择合适的系数满足等号成立的条 件. 例如,在例 1.1.2 中,使用下列形式的 AM-GM 不等式是错误的( 因为等号 不成立）x 3(1+y )(1+z )+(y +1)+(z +1)≥3x 对于每个问题,为 AM-GM 不等式给出一个固定的形式是很困难的. 这取决于你的智慧,但是寻找等号成立的条件是可以帮助我们做到这一点的. 例如， 在上面的问题中,猜测到等号成立的条件是 x =y =z =1, 为了使各个项相等,我们就选择系数 18x 3(1+y )(1+z )+y +18+z +18≥3x 4变量相等使等号成立的这样的问题,在使用 AM-GM 不等式之前是很容易确定的. 对非对称问题, 这个方法需要有一定的灵活性(参见例 1. 1. 13, 1. 1. 14 和 1. 1. 16). 有时你需要建立方程组并求出等号成立的条件(这个方法称为“平衡系数法”,这部分内容将在第 6 章中讨论).Cauchy 求反技术在本节中,我们将 AM-GM 不等式关联到一个特别的技术,称为 Cauchy 求反技术. 出乎意料的简单,而且十分有效,是这一技术的特殊优势. 下面的例子 体现了这种优势. (Bulgaria TST 2003) 设 a,b,c >0,a +b +c =3, 证明a 1+b 2+b 1+c 2+c 1+a 2≥32Proof. 事实上,直接对分母使用 AM-GM 不等式是不行的,因为不等式改变了方向,即a 1+b 2+b 1+c 2+c 1+a 2≤a 2b +b 2c +c 2a ≥32?! 但是,我们可以以另外的形式使用 AM-GM 不等式a 1+b 2=a −ab 21+b 2≥a −ab 22b =a −ab 2这样,不等式变成∑a 1+b2ccc ≥∑aosc−12∑ac,cb≥32这是由于3(∑acy b)≤(∑aecc)2=9. 当且仅当a=b=c=1时等号成立. ◻注意 使用类似的方法可以证明下列结果.设a,b,c,d是正实数, 且a+b+c+d=4, 证明a 1+b2+b1+c2+c1+d2+d1+a2≥2这个解法似乎像变魔术, AM-GM 不等式应用的两个类似的方法,带来了两个不同的解法,一个是错误的,另一个是正确的. 这种神奇出现在哪里呢? 这不足为奇,这一切来自一个分式的不同的表示形式a 1+b2=a−ab21+b2使用分式ab 21+b2前面的减号,我们可以对分母1+b2使用AM−GM不等式而不改变不等式的方向. 改变一个表达式的得号到另一个表达式,然后估计第二个表达式,是这种技术的关键所在.(Pham Kim Hung) 设a,b,c,d>0,a+b+c+d=4, 证明a1+b2c +b1+c2d+c1+d2a+d1+a2b≥2Proof.根据 AM-GM 不等式,我们有a1+b2c =a−ab2c1+b2c≥a−ab2c2b√c=a−ab√c2=a−b√a⋅ac2≥a−b(a+ac)4依据这个估计式,我们有∑a1+b2ccyc ≥∑acyc−14∑acycb−14∑acyebc再次使用 AM-GM 不等式, 很容易得到∑a ryc b≤14(∑acye)2=4,∑acycbc≤116(∑acyc)3=4因此a1+b2c +b1+c2d+c1+d2a+d1+a2b≥a+b+c+d−2=2◻设a,b,c>0, 证明a3 a2+b2+b3b2+c2+c3c2+d2+d3d2+a2≥a+b+c+d2Proof.我们有下列估计式a3 a2+b2=a−ab2a2+b2≥a−ab22ab=a−b2◻注意本题关于 4 个变量的类似结果a4a3+2b3+b4b3+2c3+c4c3+2d3+d4d3+2a4≥a+b+c+d3设a,b,c>0,a+b+c=3, 证明a2 a+2b2+b2b+2c2+c2c+2a2≥1Proof.根据 AM-GM 不等式我们有下列估计式a2 a+2b2=a−2ab2a+2b2≥a−2ab23√ab43=a−2(ab)233这就意味着∑a2 a+2b2cye ≥∑aryc−23∑(ab)23cyc于是,只需证明(ab)23+(bc)23+(ca)23≤3由 AM-GM 不等式,我们即得结果,因为(∑a cyc)2=23(∑a cyc )2+13(∑a cyc)2≥2∑a cyc +∑a cyc b=∑(a +a +ab )cye≥3∑(ab )23cyc◻ 注意 当我们把条件 a +b +c =3 改成 ab +bc +ca =3 或者 √a +√b +√c =3 时,不等式依然成立(第二个条件的情形稍微有些困难),这些问题留给读者, 这里就不给出解答了.设 a,b,c >0,a +b +c =3, 证明a 2a +2b 3+b 2b +2c 3+c 2c +2a 3≥1 Proof. 使用例 1.2.4 同样的技术,我们只需证明b √a 23+c √b 23+a √c 23≤3根据 AM-GM 不等式,我们有3∑a cyc ≥∑a cyc +2∑a cyc b =∑(a +ac +ac )cyc ≥3∑a cyc √c 23等号, 当且仅当 a =b =c =1 时成立. ◻设 a,b,c >0,a +b +c =3, 证明a +1b 2+1+b +1c 2+1+c +1a 2+1≥3 Proof. 我们使用下列估计式a +1b 2+1=a +1−b 2(a +1)b 2+1≥a +1−b 2(a +1)2b =a +1−ab +b 2关于 a,b,c 类似的结果相加,我们有∑a+1 b2+1cyc ≥3+12∑acye−12∑acycb≥3◻注意下面是四变量的类似的问题.(1) 设a,b,c,d>0,a+b+c+d=4, 证明a+1 b2+1+b+1c2+1+c+1d2+1+d+1a2+1≥4(2) 设a,b,c,d>0,a+b+c+d=4, 证明1 a2+1+1b2+1+1c2+1+1d2+1≥2设a,b,c>0,a+b+c=3, 证明11+2b2c +11+2c2a+11+2a2b≥1Proof.我们使用下列估计式11+2b2c =1−2b2c1+2b2c≥1−2√b2c33≥1−2(2b+c)9◻(Pham Kim Hung) 设a,b,c,d≥0,a+b+c+d=4, 证明1+ab 1+b2c2+1+bc1+c2d2+1+cd1+d2a2+1+da1+a2b2≥4Proof.应用 AM-GM 不等式,我们有1+ab 1+b2c2=(1+ab)−(1+ab)b2c21+b2c2≥1+ab−12(1+ab)bc类似的结果相加,我们有∑1+ab 1+b2c2cyc ≥4+∑acycb−12∑bcycc(1+ab)=4+12(∑acycb−∑acycb2c)于是,只需证明ab+bc+cd+da≤ab2c+bc2d+cd2a+da2b应用类似的结果xy+yz+zt+tx≤14(x+y+z+t)2我们得到(ab+bc+cd+da)2≥4(ab2c+bc2d+cd2a+da2b)16=(a+b+c+d)2≥4(ab+bc+cd+da)将上述不等式相乘,即得结果. 等号成立的条件是: a=b=c=d=1或者a=c= 0(b,d任意) 或者b=d=0(a,c任意). ◻(Pham Kim Hung) 设a,b,c>0,a2+b2+c2=3, 证明1 a3+2+1b3+2+1c3+2≥1Proof.根据 AM-GM 不等式,我们有∑1 a3+2cyc =32−12∑a3a3+1+1cyc≥32−12∑a33acyc=1◻。
