不等式问题

êÆc�šŽùŒ êÆ¿m¥�Ø�ª¯K o d 2012.04 Ä�Ø�ª9ÙA^ 1. �x, y, z�¢ê, …x+ y + z > xyz. ¦x 2 + y2 + z2 xyz ��Š. )‰˜ 5¿�, �x = y = z = √ 3ž, x+ y + z = xyz, x2 + y2 + z2 xyz = √ 3. e¡y² x2 + y2 + z2 xyz ��Š √ 3. ¯¢þ, k x2 + y2 + z2 > 1 3 (x+ y + z)2 >  1 3 (xyz)2 > √ 3xyz, XJxyz > 3 √ 3, 3 3 √ (xyz)2 > √ 3xyz, XJxyz < 3 √ 3. � x2 + y2 + z2 xyz ��Š √ 3. )‰� x2 + y2 + z2 > 1 3 (x + y + z)2 = 1 3 √ x+ y + z · √ (x+ y + z)3 > 1 3 √ x+ y + z √ 27xyz > 1 3 √ xyz √ 27xyz = √ 3xyz. Ø�ªÓž��Ò�^‡´x = y = z¿…x+y+z = xyz,=x = y = z = √ 3. ��x = y = z = √ 3ž, x2 + y2 + z2 xyz k�Š √ 3. )‰n x2+y2+z2 > xy+yz+zx > √ 3(xy · yz + yz · zx+ zx · xy) =√3xyz(x+ y + z) > √3xyz. �x = y = z = √ 3ž, n‡Ø�ªÓž��Ò, x2 + y2 + z2 xyz k�Š √ 3. 2. ex, y, z´�¢ê, ¦ xyz (1 + 5x)(4x+ 3y)(5y + 6z)(z + 18) �ŒŠ. )‰ 3�½y�œ¹e, x (1 + 5x)(4x+ 3y) = x 20x2 + (15y + 4)x+ 3y = 1 20x+ 3y x + 15y + 4 6 1 2 √ 20× 3y + 15y + 4 = 1 ( √ 15y + 2)2 . �…=�x = √ 3y 20 ž, �Ò¤á. ÓnŒ�, z (5y + 6z)(z + 18) 6 1 2 √ 6× 90y + 5y + 108 = 1 ( √ 5y + 6 √ 3)2 . �…=�z = √ 15yž, �Ò¤á. 1 ¤±, xyz (1 + 5x)(4x+ 3y)(5y + 6z)(z + 18) 6 y ( √ 15y + 2)2( √ 5y + 6 √ 3)2 = [ √ y ( √ 15y + 2)( √ 5y + 6 √ 3) ]2 =  1 5 √ 3y + 12 √ 3√ y + 20 √ 5  2 6 [ 1 2 √ 5 √ 3× 12√3 + 20√5 ]2 = ( 1 32 √ 5 )2 = 1 5120 , �…=�x = 3 5 , y = 12 5 , z = 6ž, þª��ŒŠ 1 5120 . 3. �x, y, z, w´Ø�"�¢ê, ¦ xy + 2yz + zw x2 + y2 + z2 + w2 �ŒŠ. )‰ Ú\�ëêα, β, γ, Kkα2x2 + y2 > 2αxy, β2y2 + z2 = 2βyz, γ2z2 + w2 > 2γzw. ò±þn ªƒ\=� α 2 x2 + ( 1 2α + β ) y2 + ( 1 β + γ 2 ) z2 + w2 2γ > xy + 2yz + zw. (1) - α 2 = 1 2α + β = 1 β + γ 2 = 1 2γ . �α = √ 2 + 1, β = 1, γ = √ 2− 1, “\ª(1) =�(Ø. 4. �x1, x2, · · · , xn ∈ R+, ½Â Sn = n∑ i=1 ( xi + n− 1 n2 · 1 xi )2 . (1)¦Sn��Š. (2)3x21 + x 2 2 + · · ·+ x2n = 1�^‡e, ¦Sn��Š. (3)3x1 + x2 + · · ·+ xn = 1�^‡e, ¦Sn��Š. )‰ (1) Sn > n∑ i=1 ( 2 √ n− 1 n )2 = 4 n∑ i=1 n− 1 n2 = 4(n− 1) n . �x1 = x2 = · · · = xn = √ n− 1 n ž, ���Š 4(n− 1) n . (2) Sn = n∑ i=1 ( x2i + 2 · n− 1 n2 + (n− 1)2 n4 · 1 x2i ) = 1 + 2 · n− 1 n + (n− 1)2 n4 · n∑ i=1 1 x2i > 1 + 2 · n− 1 n + (n− 1)2 n2 = ( 1 + n− 1 n )2 . �x1 = x2 = · · · = xn = 1√ n ž, ���Š ( 1 + n− 1 n )2 . (3) Ϗ [ n∑ i=1 1 · ( xi + n− 1 n · 1 xi )]2 6 ( n∑ i=1 12 ) · n∑ i=1 ( xi + n− 1 n2 · 1 xi )2 , 2 ¤±, Sn = n∑ i=1 ( xi + n− 1 n2 · 1 xi )2 > 1 n [ n∑ i=1 ( xi + n− 1 n2 · 1 xi )]2 > 1 n [ 1 + n− 1 n2 · n2 ]2 = n. �x1 = x2 = · · · = xn = 1 n ž, ���Šn. 5. ��¢êa, b, c÷vabc = 1. ¦y: éu�êk > 2, k ak a+ b + bk b+ c + ck c+ a > 3 2 . )‰ Ϗ ak a+ b + 1 4 (a+ b) + 1 2 + 1 2 + · · ·+ 1 2︸ ︷︷ ︸ k−2‡ > k k √ ak 2k = k 2 a, ¤± ak a+ b > k 2 a− 1 4 (a+ b)− k − 2 2 . ÓnŒ�, bk b+ c > k 2 b− 1 4 (b+ c)− k − 2 2 , ck c+ a > k 2 c− 1 4 (c+ a)− k − 2 2 . nªƒ\Œ�, ak a+ b + bk b+ c + ck c+ a > k 2 (a+ b+ c)− 1 2 (a+ b+ c)− 3 2 (k − 2) = k − 1 2 (a+ b+ c)− 3 2 (k − 2) > 3 2 (k − 1)− 3 2 (k − 2) = 3 2 . 6. �a, b, c´�¢ê, ¦y: √ abc (√ a+ √ b+ √ c ) + (a+ b+ c)2 > 4 √ 3abc(a+ b+ c). )‰˜ ؔ�a+ b+ c = 1, K�Ø�ªŒz √ abc (√ a+ √ b+ √ c ) + 1 > 4 √ 3abc, = √ a+ √ b+ √ c+ 1√ abc > 4 √ 3. dþŠØ�ªŒ� √ a+ √ b+ √ c+ 1√ abc = √ a+ √ b+ √ c+ 1 9 √ abc + 1 9 √ abc + · · ·+ 1 9 √ abc > 12 ·  √abc( 9 √ abc )9  1 12 = 12 · 1 9 3 4 (abc) 1 3 = 4√ 3 · 1 (abc) 1 3 > 4√ 3 · 1 a+ b+ c 3 = 4 √ 3. )‰� ؔ�abc = 1, K √ a+ √ b+ √ c > √ 3, a+ b+ c > 3, �Ø�ªŒz (√ a+ √ b+ √ c ) + (a+ b+ c)2 > 4 √ 3(a+ b+ c), = √ a+ √ b+ √ c√ a+ b+ c + (√ a+ b+ c )3 > 4 √ 3. 3 dþŠØ�ª, � √ a+ √ b+ √ c√ a+ b+ c + (√ a+ b+ c )3 = √ a+ √ b+ √ c√ a+ b+ c + (√ a+ b+ c )3 3 + (√ a+ b+ c )3 3 + (√ a+ b+ c )3 3 > 4 4 √(√ a+ √ b+ √ c ) · 1 33 (a+ b+ c)4 > 4 4 √ 32 = 4 √ 3. 7. �x, y, z´Œu−1�¢ê. y²: 1 + x2 1 + y + z2 + 1 + y2 1 + z + x2 + 1 + z2 1 + x+ y2 > 2. )‰˜ d®�1 + x2, 1 + y2, 1 + z2, 1 + y + z2, 1 + z + x2, 1 + x+ y2þŒu0. d…ÜØ�ª, �( 1 + x2 1 + y + z2 + 1 + y2 1 + z + x2 + 1 + z2 1 + x+ y2 ) [(1 + x2)(1 + y + z2) + (1 + y2)(1 + z + x2) + (1 + z2)(1 + x+ y2)] > (1 + x2 + 1 + y2 + 1 + z2)2. l 1 + x2 1 + y + z2 + 1 + y2 1 + z + x2 + 1 + z2 1 + x+ y2 > (x 2 + y2 + z2 + 3)2 (1 + x2)(1 + y + z2) + (1 + y2)(1 + z + x2) + (1 + z2)(1 + x+ y2) = x4 + y4 + z4 + 9 + 2x2y2 + 2y2z2 + 2z2x2 + 6x2 + 6y2 + 6z2 x2y2 + y2z2 + z2x2 + 2(x2 + y2 + z2) + x2y + y2z + z2x+ x+ y + z + 3 = 2 + x4 + y4 + z4 + 3 + 2x2 + 2y2 + 2z2 − 2(x2y + y2z + z2x) x2y2 + y2z2 + z2x2 + 2(x2 + y2 + z2) + x2y + y2z + z2x+ x+ y + z + 3 = 2 + (x2 − y)2 + (y2 − z)2 + (z2 − x)2 + (x− 1)2 + (y − 1)2 + (z − 1)2 x2y2 + y2z2 + z2x2 + 2(x2 + y2 + z2) + x2y + y2z + z2x+ x+ y + z + 3 > 2. �…=�x = y = z = 1ž, þª�Ò¤á. )‰� d®�1 + x2, 1 + y2, 1 + z2, 1 + y + z2, 1 + z + x2, 1 + x+ y2þŒu0, u´ 1 + x2 1 + y + z2 + 1 + y2 1 + z + x2 + 1 + z2 1 + x+ y2 > 1 + x 2 1 + z2 + 1 + y2 2 + 1 + y2 1 + x2 + 1 + z2 2 + 1 + z2 1 + y2 + 1 + x2 2 = 2a 2c+ b + 2b 2a+ c + 2c 2b+ a , Ù¥, a = 1 + x2 2 , b = 1 + y2 2 , c = 1 + z2 2 . d…ÜØ�ª, � 2a 2c+ b + 2b 2a+ c + 2c 2b+ a > (a+ b+ c) 2 a(b+ 2c) + b(c+ 2a) + c(a+ 2b) > 3(ab+ bc+ ca) 3(ab+ bc+ ca) = 1. 8. �a, b, c > 0. y²: (a− b)2 (c+ a)(c+ b) + (b− c)2 (a+ b)(a+ c) + (c− a)2 (b+ c)(b+ a) > (a− b) 2 a2 + b2 + c2 . 4 )‰˜ d 1 2 (a − 2b)2 + 1 2 (a − 2c)2 + (b − c)2 > 0 ⇒ 3(a2 + b2 + c2) > 2a2 + 2ab + 2bc + 2ac = 2(a+ b)(a+ c)⇒ (a+ b)(a+ c) 6 3 2 (a2 + b2 + c2). Ón, (b+ a)(b+ c) 6 3 2 (a2 + b2 + c2), (c+ a)(c+ b) 6 3 2 (a2 + b2 + c2). � (a− b)2 (c+ a)(c+ b) + (b− c)2 (a+ b)(a+ c) + (c− a)2 (b+ c)(b+ a) > 2 3 · (a− b) 2 + (b− c)2 + (c− a)2 a2 + b2 + c2 > 2 3 · (a− b)2 + 1 2 (b− c+ c− a)2 a2 + b2 + c2 = (a− b)2 a2 + b2 + c2 . )‰� P“ ∑ ” L«=†é¡Ú. d…ÜØ�ª ∑ (a− b)2 (c+ a)(c+ b) · ∑ (c+ a)(c+ b) > (|a− b|+ |b− c|+ |c− a|)2 > (|a− b|+ |b− c+ c− a|)2 = 4(a− b)2. ∑ (c+ a)(c+ b) = ∑ a2 + 3 ∑ ab 6 4 ∑ a2, � ∑ (a− b)2 (c+ a)(c+ b) > 4(a− b) 2 4(a2 + b2 + c2) = (a− b)2 a2 + b2 + c2 . 9. ��¢êa1, a2, · · · , an÷va1 + a2 + · · ·+ an = 1. ¦y: (a1a2 + a2a3 + · · ·+ ana1) ( a1 a22 + a2 + a2 a23 + a3 + · · ·+ an a21 + a1 ) > n n+ 1 . )‰˜ Äkd…ÜØ�ª´�eãÚn: Ún �a1, a2, · · · , an´¢ê, x1, x2, · · · , xn´�ê, K a21 x1 + a22 x2 + · · ·+ a 2 n xn > (a1 + a2 + · · ·+ an) 2 x1 + x2 + · · ·+ xn . dÚn9K�� a1 a2 + a2 a3 + · · ·+ an a1 = a21 a1a2 + a22 a2a3 + · · ·+ a 2 n ana1 > 1 a1a2 + a2a3 + · · ·+ ana1 . Ï Ly² a1 a22 + a2 + a2 a23 + a3 + · · ·+ an a21 + a1 > n n+ 1 ( a1 a2 + a2 a3 + · · ·+ an a1 ) . (1) dÚn� a1 a22 + a2 + a2 a23 + a3 + · · ·+ an a21 + a1 = ( a1 a2 )2 a1 + a1 a2 + ( a2 a3 )2 a2 + a2 a3 + · · ·+ ( an a1 )2 an + an a1 > ( a1 a2 + a2 a3 + · · ·+ an a1 )2 1 + a1 a2 + a2 a3 + · · ·+ an a1 . (2) -t = a1 a2 + a2 a3 + · · ·+ an a1 , Kt > n. l Ly t2 1 + t > nt n+ 1 , 5 dª�dut > n, y.. )‰� ½Âan+1 = a1. e n∑ i=1 aiai+1 > 1 n , K n∑ i=1 ai ai+1(ai+1 + 1) n∑ i=1 (ai+1 + 1) > ( n∑ i=1 ( ai ai+1 ) 1 2 )2 > n, u´, n∑ i=1 ai ai+1(ai+1 + 1) > n 2 n+ 1 , l , n∑ i=1 aiai+1 n∑ i=1 ai ai+1(ai+1 + 1) > n n+ 1 . e n∑ i=1 aiai+1 < 1 n , K ( n∑ i=1 aiai+1 )( n∑ i=1 ai ai+1(ai+1 + 1) )( n∑ i=1 ai(ai+1 + 1) ) > ( n∑ i=1 ai )3 = 1, q n∑ i=1 ai(ai+1 + 1) < n+ 1 n , ·‚k n∑ i=1 aiai+1 n∑ i=1 ai ai+1(ai+1 + 1) > n n+ 1 . 10. �a, b, c, d�¢ê,÷vab+ cd = 1,:Pi(xi, yi) (i = 1, 2, 3, 4)´±�:�%�ü �±þ� o‡:. ¦y: (ay1 + by2 + cy3 + dy4) 2 + (ax4 + bx3 + cx2 + dx1) 2 6 2 ( a2 + b2 ab + c2 + d2 cd ) . )‰˜ -u = ay1 + by2, v = cy3 + dy4, u1 = ax4 + bx3, v1 = cx2 + dx1, K u2 6 (ay1 + by2)2 + (ax1 − bx2)2 = a2 + b2 + 2ab(y1y2 − x1x2), = x1x2 − y1y2 6 a 2 + b2 − u2 2ab . (1) v21 6 (cx2 + dx1)2 + (cy2 − dy1)2 = c2 + d2 + 2cd(x1x2 − y1y2), = y1y2 − x1x2 6 c 2 + d2 − v21 2cd . (2) (1) + (2), � 0 6 a 2 + b2 − u2 ab + c2 + d2 − v2 cd , = u2 ab + v21 cd 6 a 2 + b2 ab + c2 + d2 cd . Ón, v2 cd + u21 ab 6 c 2 + d2 cd + a2 + b2 ab . 6 d…ÜØ�ª, k (u+ v)2 + (u1 + v1) 2 6 (ab+ cd) ( u2 ab + v2 cd ) + (ab+ cd) ( u21 ab + v21 cd ) = u2 ab + v2 cd + u21 ab + v21 cd 6 2 ( a2 + b2 ab + c2 + d2 cd ) . )‰� Pα = ay1 + by2 + cy3 + dy4, β = ax4 + bx3 + cx2 + dx1. d…ÜØ�ª, � [( √ ady1) 2 + ( √ bcy2) 2 + ( √ bcy3) 2 + ( √ ady4) 2] · (√a d )2 + (√ b c )2 + (√ c b )2 + (√ d a )2 > (ay1 + by2 + cy3 + dy4)2 = α2, = α2 6 (ady21 + bcy22 + bcy23 + ady24) · ( a d + b c + c b + b a ) . Ón, β2 6 (adx24 + bcx23 + bcx22 + adx21) · ( a d + b c + c b + b a ) . ò±þüªƒ\, ¿|^x2i + y 2 i = 1 (i = 1, 2, 3, 4) , ab+ cd = 1�, α2 + β2 6 (2ad+ 2bc) ( a d + b c + c b + b a ) = 2(ad+ bc) ( ab+ cd bd + ab+ cd ac ) = 2(ad+ bc) ( 1 bd + 1 ac ) = 2 ( a2 + b2 ab + c2 + d2 cd ) . nܯKÀù 11. ¦¤k�¢êk, ¦�Ø�ª a3 + b3 + c3 + d3 + 1 > k(a+ b+ c+ d) é?¿a, b, c, d ∈ [−1,+∞)Ѥá. )‰�a = b = c = d = −1ž,k−3 > k ·(−4),=k > 3 4 . �a = b = c = d = 1 2 ž,k4 · 1 8 +1 > k ·2, =k 6 3 4 . l k = 3 4 . e¡y²Ø�ª a3 + b3 + c3 + d3 + 1 > 3 4 (a+ b+ c+ d) (1) é?¿a, b, c, d ∈ [−1,+∞)Ѥá. Äky²4x3+1 > 3x, x ∈ [−1,+∞). ¯¢þ,�x ∈ [−1,+∞)ž, 4x3+1−3x = (x+1)(2x−1)2 > 0. l k 4a3 + 1 > 3a, 4b3 + 1 > 3b, 4c3 + 1 > 3c, 4d3 + 1 > 3d, ò±þo‡Ø�ªƒ\, =�ª(1) . Ïd, ¤¦�¢êk = 3 4 . 7 12. ®a, b, c´�¢ê. ¦y: (2a+ b+ c)2 2a2 + (b+ c)2 + (2b+ c+ a)2 2b2 + (c+ a)2 + (2c+ a+ b)2 2c2 + (a+ b)2 6 8. )‰˜éa, b, c¦˜‡Ü·�ÏfŒr�¯Ky8a+ b+ c = 3 (a, b, c > 0)�œ/, �Ly ² (a+ 3)2 2a2 + (3− a)2 + (b+ 3)2 2b2 + (3− b)2 + (c+ 3)2 2c2 + (3− c)2 6 8. -f(x) = (x+ 3)2 2x2 + (3− x)2 . Ly²f(a) + f(b) + f(c) 6 8. 5¿� f(x) = x2 + 6x+ 9 3(x2 − 2x+ 3) = 1 3 · x 2 + 6x+ 9 x2 − 2x+ 3 = 1 3 ( 1 + 8x+ 6 x2 − 2x+ 3 ) = 1 3 ( 1 + 8x+ 6 (x− 1)2 + 2 ) 6 1 3 ( 1 + 8x+ 6 2 ) = 1 3 (4x+ 4), Kf(a) + f(b) + f(c) 6 1 3 (4a+ 4 + 4b+ 4 + 4c+ 4) = 8. )‰� d…ÜØ�ª� [a2 + a2 + (b+ c)2](12 + 12 + 22) > [a+ a+ 2(b+ c)]2, =2a2 + (b+ c)2 > 2 3 (a+ b+ c)2, l , 1 2a2 + (b+ c)2 6 3 2(a+ b+ c)2 . Ón, 1 2b2 + (c+ a)2 6 3 2(a+ b+ c)2 , 1 2c2 + (a+ b)2 6 3 2(a+ b+ c)2 . 5¿� (2a+ b+ c)2 2a2 + (b+ c)2 + (2b+ c+ a)2 2b2 + (c+ a)2 + (2c+ a+ b)2 2c2 + (a+ b)2 − 8 = [ (2a+ b+ c)2 2a2 + (b+ c)2 − 1 ] + [ (a+ 2b+ c)2 2b2 + (c+ a)2 − 1 ] + [ (a+ b+ 2c)2 2c2 + (a+ b)2 − 1 ] − 5 = 2a2 + 4ab+ 4ac 2a2 + (b+ c)2 + 2b2 + 4ab+ 4bc 2b2 + (c+ a)2 + 2c2 + 4ac+ 4bc 2c2 + (a+ b)2 − 5 6 3[(2a 2 + 4ab+ 4ac) + (2b2 + 4ab+ 4bc) + (2c2 + 4ac+ 4bc)] 2(a+ b+ c)2 − 5 = −2[(a− b)2 + (b− c)2 + (c− a)2] (a+ b+ c)2 6 0. ��Ø�ª¤á. 13. �a, b, c�¢ê, ¦ a+ 3c a+ 2b+ c + 4b a+ b+ 2c − 8c a+ b+ 3c ��Š. )‰˜-x = a+2b+ c, y = a+ b+2c, z = a+ b+3c, ddŒ)�a+3c = 2y− x, b = z+ x− 2y, c = z − y, l , a+ 3c a+ 2b+ c + 4b a+ b+ 2c − 8c a+ b+ 3c = 2y − x x + 4(z + x− 2y) y − 8(z − y) z = −17 + 2y x + 4 x y + 4 z y + 8 y z > −17 + 2 √ 8 + 2 √ 32 = −17 + 12 √ 2 8 ª¥�Ò�…=�y = √ 2x, z = 2x=b = (1 + √ 2)a, c = (4 + 3 √ 2)až¤á. Ïd¤¦��Š −17 + 12√2. )‰� ؔ�a+ b+ c = 1, K a+ 3c a+ 2b+ c + 4b a+ b+ 2c − 8c a+ b+ 3c = 1 + 2c− b 1 + b + 4b 1 + c − 8c 1 + 2c = −1 + 2 + 2c 1 + b + 4b+ 4 1 + c − 4 1 + c + 4 1 + 2c − 4 = −5 + 21 + c 1 + b + 4 1 + b 1 + c − 4c (1 + c)(1 + 2c) > −5 + 2 √ 8 = 4 1 c + 3 + 2c > −5 + 4 √ 2− 4 3 + 2 √ 2 = −17 + 12 √ 2. �a = 3− 2√2 2 , b = √ 2− 1 2 , c = √ 2 2 ž, �Ò¤á. Ïd¤¦��Š−17 + 12√2. 14. �¢êx, y, z÷vxyz > 1. y²: x5 − x2 x5 + y2 + z2 + y5 − y2 y5 + z2 + x2 + z5 − z2 z5 + x2 + y2 > 0. )‰˜ �Ø�ªŒC/ x2 + y2 + z2 x5 + y2 + z2 + x2 + y2 + z2 y5 + z2 + x2 + x2 + y2 + z2 z5 + x2 + y2 6 3. d…ÜØ�ª9K�^‡xyz > 1, � (x5 + y2 + z2)(yz + y2 + z2) > [x2(xyz) 12 + y2 + z2]2 > (x2 + y2 + z2)2. = x2 + y2 + z2 x5 + y2 + z2 6 yz + y 2 + z2 x2 + y2 + z2 . Ón, x2 + y2 + z2 y5 + z2 + x2 6 zx+ z 2 + x2 x2 + y2 + z2 , x2 + y2 + z2 z5 + x2 + y2 6 xy + x 2 + y2 x2 + y2 + z2 . rþ¡n‡Ø�ªƒ\, ¿|^x2 + y2 + z2 > xy + yz + zx, � x2 + y2 + z2 x5 + y2 + z2 + x2 + y2 + z2 y5 + z2 + x2 + x2 + y2 + z2 z5 + x2 + y2 6 2 + xy + yz + zx x2 + y2 + z2 6 3. )‰� Ϗ x5 − x2 x5 + y2 + z2 − x 5 − x2 x3(x2 + y2 + z2) = x2(x3 − 1)2(y2 + z2) x3(x5 + y2 + z2)(x2 + y2 + z2) > 0, ¤±,∑ x5 − x2 x5 + y2 + z2 > ∑ x5 − x2 x3(x2 + y2 + z2) = 1 x2 + y2 + z2 ∑( x2 − 1 x ) > 1 x2 + y2 + z2 ∑ (x2− yz) > 0. 15. �xi > 0 (i = 1, 2, · · · , n) , … n∑ i=1 x2i + 2 ∑ 16k 1, 9 � n∑ i=1 xi > 1, �Ò¤á�…=�3i¦�xi = 1, xj = 0, j 6= i. Ïd, n∑ i=1 xi��Š1. 2¦ŒŠ. -xk = √ kyk, k n∑ k=1 ky2k + 2 ∑ 16k a2 > · · · > an, l , yk = ak − ak−1 = 2 √ k − (√k + 1 +√k − 1)[ n∑ k=1 ( √ k −√k − 1)2 ] 1 2 > 0, =xk > 0. ¤¦ŒŠ [ n∑ k=1 ( √ k −√k − 1)2 ] 1 2 . 16. �x, y, z´�¢ê, …÷vx+ y + z = 1. ¦y: xy√ xy + yz + yz√ yz + zx + zx√ zx+ xy 6 √ 2 2 . )‰˜ ·‚y²˜‡r�Ø�ª: xy√ xy + yz + yz√ yz + zx + zx√ zx+ xy 6 3 √ 3 4 √ (x+ y)(y + z)(z + x). (1) 10 w,, ª(1) �du f = √ x (z + x)(x+ y) · xy (y + z)(z + x) + √ y (x+ y)(y + z) · yz (z + x)(x+ y) + √ z (y + z)(z + x) · zx (x+ y)(y + z) 6 3 √ 3 4 . duf'ux, y, zӆé¡, ؔ�x = min{x, y, z}. L©x 6 y 6 zÚx 6 z 6 yü«œ¹y². du ü«œ¹�y²�Ÿþ��ƒÓ, �=y1˜«œ¹. dx 6 y 6 z ⇒ xy 6 zx 6 yz, (y + z)(z + x) > (y + z)(x+ y) > (x+ y)(z + x), � xy (y + z)(z + x) 6 zx (x+ y)(y + z) 6 yz (z + x)(x+ y) . (2) qx(y + z) 6 y(z + x) 6 z(x+ y), � x (z + x)(x+ y) 6 y (x+ y)(y + z) 6 z (y + z)(z + x) . (3) dª(2) (3) 9üSØ�ª f 6 √ x2y (x+ y)(z + x)2(y + z) + √ xyz (x+ y)2(y + z)2 + √ yz2 (z + x)2(x+ y)(y + z) = √ xyz (x+ y)2(y + z)2 + 2× 1 2 √ y (x+ y)(y + z) 6 √ 3 [ xyz (x+ y)2(y + z)2 + 2× 1 4 y (x+ y)(y + z) ] . Ïd, ‡yf 6 3 √ 3 4 , Ly xyz (x+ y)2(y + z)2 + 1 2 · y (x+ y)(y + z) 6 9 16 ⇔ 16xyz + 8y(x+ y)(y + z) 6 9(x+ y)2(y + z)2 ⇔ 9x2z2 + y2 > 6xyz ⇔ (3xz − y)2 > 0. Ïd, Ø�ª(2) ¤á. d(x+ y)(y + z)(z + x) 6 [ 2(x+ y + z) 3 ]3 = 8 27 , “\ª(1) =��Ø�ª. )‰� -x = a2, y = b2, z = c2, Ka2 + b2 + c2 = 1. u´, �Ø�ª�du∑ a2b2√ a2b2 + b2c2 6 √ 2 2 . Ϗ √ a2b2 + b2c2 > √ 2 2 (ab+ bc), �LyA = ∑ a2b2 ab+ bc 6 1 2 . �EéóªB = ∑ b2c2 ab+ bc , K A−B = ∑ ab2 − bc2 ab+ bc = ∑ (ab− bc) = 0. Ïd, A = B. �A 6 1 2 ⇔ B 6 1⇔ ∑ a2b2 + b2c2 ab+ bc 6 1⇔ ∑( b · a 2 + c2 a+ c ) 6 ∑ b2. ∑( b · a 2 + c2 a+ c ) − ∑ b2 = ∑( b · a 2 + c2 a+ c − b2 ) = ∑ b a+ c [a(a− b) + c(c− b)] = ∑ ab(a− b) a+ c + ∑ bc(c− b) a+ c = ∑ ab(a− b) a+ c + ∑ bc(c− b) c+ b = ∑[ ab(a− b) ( 1 a+ c − 1 b+ c )] = − ∑ ab(a− b)2 (a+ c)(b+ c) 6 0. 11 Ïd, �Ø�ª¤á. )‰n 5¿� 2x x+ z + 9xy > 6 √ 2 · x √ y√ x+ z = 6 √ 2xy√ xy + zy , K xy√ xy + zy 6 1 6 √ 2 ( 2x x+ z + 9xy ) . u´, Ly ∑ 2x x+ z + 9 ∑ xy 6 √ 2 2 × 6 √ 2 = 6, =9 ∑ xy 6 ∑ 2z x+ z . d…ÜØ�ªk ∑ z x+ z · ∑ z(x+ z) > (∑ x )2 = 1, K ∑ z x+ z > 1 x2 + y2 + z2 + xy + yz + xz = 1(∑ x )2 − ∑ xy . ¤±, Ly 9 ∑ xy 6 2(∑ x ) − ∑ xy , =9b 6 2 a2 − b , Ù¥, a = x+ y + z, b = xy + yz + zx. Ϗ9b(a2 − b) 6 2 = 2a4 ⇔ (a2 − 3b)(2a2 − 3b) > 0, a2 > 3b. ¤±, �Ø�ª¤á. )‰o d…ÜØ�ªk(∑ xy√ xy + yz )2 6 (∑ xy xy + yz )(∑ xy ) = ∑( xy + 2 x−1 + y−1 · x 2 ) 6 ∑( xy + x(x+ z) 4 ) = 1 4 + 3 4 ∑ xy 6 1 4 + 1 4 (x+ y + z)2 = 1 2 . ¤±, �Ø�ª¤á. 17. ®n (n > 2) ‡�¢êa1, a2, · · · , an÷v n∑ i=1 ai · n∑ i=1 1 ai 6 ( n+ 1 2 )2 . ¦y: max{a1, a2, · · · , an} 6 4min{a1, a2, · · · , an}. )‰ Pm = min{a1, a2, · · · , an}, M = max{a1, a2, · · · , an}. Šâé¡5, ؔ�m = a1 6 a2 6 · · · 6 an =M . 5¿�( n+ 1 2 )2 > (a1 + a2 + · · ·+ an) ( 1 a1 + 1 a2 + · · ·+ 1 an ) = (m+ a2 + · · ·+ an−1 +M) · ( 1 M + 1 a2 + · · ·+ 1 an−1 + 1 m ) > √m M + 1 + · · ·+ 1︸ ︷︷ ︸ n−2‡ + √ M m  2 . Kkn+ 1 2 > √ m M + n− 2 + √ M m ⇒ 2(m+M) 6 5 √ Mm⇒M 6 4m. 12 18. ‰½Œu3��ên, �¢êx1, x2, · · · , xn, xn+1, xn+2÷v^‡0 < x1 < x2 < · · · < xn < xn+1 < xn+2. Á¦ ( n∑ i=1 xi+1 xi ) n∑ j=1 xj+2 xj+1  ( n∑ k=1 xk+1xk+2 x2k+1 + xkxk+2 )( n∑ l=1 x2l+1 + xlxl+2 xlxl+1 ) ��Š, ¿¦Ñ¦Tªˆ��Š�¤k÷v^‡�¢ê|x1, x2, · · · , xn, xn+1, xn+2. )‰ (I) Pti = xi+1 xi (> 1) , 1 6 i 6 n+ 1. K¥�ªfŒ�¤ ( n∑ i=1 ti )( n∑ i=1 ti+1 ) ( n∑ i=1 titi+1 ti + ti+1 )( n∑ i=1 (ti + ti+1) ) . ·‚w�, ( n∑ i=1 titi+1 ti + ti+1 )( n∑ i=1 (ti + ti+1) ) = ( n∑ i=1 ti − n∑ i=1 t2i ti + ti+1 )( n∑ i=1 (ti + ti+1) ) = ( n∑ i=1 ti )( n∑ i=1 (ti + ti+1) ) − ( n∑ i=1 t2i ti + ti+1 )( n∑ i=1 (ti + ti+1) ) 6 ( n∑ i=1 ti )( n∑ i=1 (ti + ti+1) ) − ( n∑ i=1 ti√ ti + ti+1 √ ti + ti+1 )2 = ( n∑ i=1 ti )2 + ( n∑ i=1 ti )( n∑ i=1 ti+1 ) − ( n∑ i=1 ti )2 = ( n∑ i=1 ti )( n∑ i=1 ti+1 ) . Ïd, éÎÜ^‡�¢ê|0 < x1 < x2 < · · · < xn < xn+1 < xn+2, K¥�ªfØ�u1. (II) þ¡�íü^�…ÜØ�ª, �Ò¤á�¿©7‡^‡´ √ ti + ti+1 ti√ ti + ti+1 = d, 1 6 i 6 n, ùp, d~ê. Ò´ ti+1 ti = d − 1 = c, 1 6 i 6 n. Pt1 = b, ktj = bcj−1, 1 6 j 6 n + 1. ƒA/ k xj+1 xj = tj = bc j−1, 1 6 j 6 n+ 1. Px1 = a > 0, k xk = tk−1tk−2 · · · t1a = abk−1c (k−1)(k−2) 2 , 2 6 k 6 n+ 2. dx2 > x1, �b = x2 x1 > 1. qtj = bcj−1 > 1, 1 6 j 6 n+ 1, l , c > n √ 1 b (> j−1 √ 1 b , 1 6 j 6 n+ 1) . (III) ��(Ø: (i) éuÎÜ^‡�¢ê|x1, x2, · · · , xn, xn+1, xn+2, K¥ªf��Š´1. 13 (ii)U¦Tªˆ��Š�ÎÜ^‡0 < x1 < x2 < · · · < xn < xn+1 < xn+2�¢ê|x1, x2, · · · , xn, xn+1, xn+2A T´x1 = a, xk = abk−1c (k−1)(k−2) 2 , 2 6 k 6 n+ 2, Ù¥a > 0, b > 1, c > n √ 1 b . 19. �f(x, y, z) = x(2y − z) 1 + x+ 3y + y(2z − x) 1 + y + 3z + z(2x− y) 1 + z + 3x ,Ù¥x, y, z > 0,…x+y+z = 1. ¦f(x, y, z)� ŒŠÚ�Š. )‰˜ kyf 6 1 7 , �…=�x = y = z = 1 3 ž�Ò¤á. Ϗ f = ∑ x(x+ 3y − 1) 1 + x+ 3y = 1− 2 ∑ x 1 + x+ 3y , (1) d…ÜØ�ª ∑ x 1 + x+ 3y > (∑ x )2 ∑ x(1 + x+ 3y) = 1∑ x(1 + x+ 3y) , q ∑ x(1 + x+ 3y) = ∑ x(2x+ 4y + z) = 2 + ∑ xy 6 7 3 . l , ∑ x 1 + x+ 3y > 3 7 , f 6 1− 2× 3 7 = 1 7 , fmax = 1 7 , �…=�x = y = z = 1 3 ž�Ò¤á. 2yf > 0, �x = 1, y = z = 0ž�Ò¤á. ¯¢þ, f(x, y, z) = x(2y − z) 1 + x+ 3y + y(2z − x) 1 + y + 3z + z(2x− y) 1 + z + 3x = xy ( 2 1 + x+ 3y − 1 1 + y + 3z ) + yz ( 2 1 + y + 3z − 1 1 + z + 3x ) + zx ( 2 1 + z + 3x − 1 1 + z + 3y ) = 7xyz (1 + x+ 3y)(1 + y + 3z) + 7xyz (1 + y + 3z)(1 + z + 3x) + 7xyz (1 + z + 3x)(1 + x+ 3y) > 0. �fmin = 0, �x = 1, y = z = 0ž�Ò¤á. )‰� ŒŠÓ)‰˜. �z = minx, y, z, ez = 0, K f(x, y, 0) = 2xy 1 + x+ 3y − xy 1 + y = 2xy 2x+ 4y − xy x+ 2y = 0. e�x, y > z > 0, d(1) ª, ‡yf > 0, ‡y∑ x 1 + x+ 3y 6 1 2 . (2) 5¿� 1 2 = x 2x+ 4y + y x+ 2y , u´(2) �du z 1 + z + 3x 6 ( x 2x+ 4y − x 1 + x+ 3y ) + ( y x+ 2y − y 1 + y + 3z ) = z 2x+ 4y ( x 1 + x+ 3y + 8y 1 + y + 3z ) , = 2x+ 4y 1 + z + 3x 6 x 1 + x+ 3y + 8y 1 + y + 3z . (3) 14 d…ÜØ�ª, Œ� x 1 + x+ 3y + 8y 1 + y + 3z = x2 x(1 + x+ 3y) + (2y)2 y(1 + y + 3z) 2 > (x+ 3y) 2 (x+ x2 + 3xy) + y + y2 + 3yz 2 = 2x+ 4y 1 + z + 3x , =(3) ¤á, l f > 0, �fmin = 0, �x = 1, y = z = 0ž�Ò¤á. 20. �a1, a2, · · · , an (n > 3) ´¢ê. y²: n∑ i−1 a2i − n∑ i=1 aiai+1 6 [n 2 ] (M −m)2, Ù¥, [x]L«Ø‡L¢êx�Œ�ê, …an+1 = a1, M = max 16i6n ai, m = min 16i6n ai. )‰ en = 2k (k ∈ N∗) , K 2 ( n∑ i=1 a2i − n∑ i=1 aiai+1 ) = n∑ i=1 (ai − ai+1)2 6 n(M −m)2. � n∑ i=1 a2i − n∑ i=1 aiai+1 6 n 2 (M −m)2 = [n 2 ] (M −m)2. en = 2k+1 (k ∈ N∗) ,Kéûü��2k+1‡ê,7këYn‘4O½4~. ÄÙ�Ï,du 2k+1∏ i=1 (ai − ai−1)(ai+1 − ai) = 2k+1∏ i=1 (ai − ai−1)2 > 0, u´, ،Uéuz˜‡i, Ñkai −