# 竞赛中各类不等式题选析

n

i =1

n

xi = 1,

∑x
i =1

n

i

=0

i =1

xi 1 1 ≤ i 2 2n

x k l +1 , x kl + 2 , , x k n ,则由已知条件得

∑ x ki =
i =1

l

1 n 1 , ∑ x ki = 2 i =l +1 2

l x n x l xi ki 1 n k =∑ i ∑ ≤ ∑ x ki ∑ x ki ∑ i i =1 k i =l +1 k i =1 n i =l +1 i =1 i i n

1 1 2 2n

i =1

n

l x n x n xi ki 1 l k =∑ i ∑ ≥ ∑ x ki ∑ x ki i n i =1 i =1 k i i = l +1 k i i = l +1

1 1 ≥ 2 2n

i =1

n

xi 1 1 ≤ i 2 2n

∑ xi = 0, ∑ xi = 1
i =1 i =1

n

n

1 , i = 1,2, , n 1 2

i =1

n

n n n 1 n 1 xi S S 1 1 1 = ∑ (S i S i 1 ) = ∑ i ∑ i = ∑ S i i i i + 1 i =1 i i =1 i i =1 i + 1 i =1

x i n 1 1 1 ∑ i ≤ ∑ Si i i + 1 i =1 i =1
n

1 n 1 1 1 1 ∑ i i + 1 = 2 1 2 i =1

1 n

2 2
2

2

≤ 2

x 2 + 2 xy y 2 = a 2 cos 2 θ + 2a 2 cos θ sin θ a 2 sin 2 θ = a 2 cos 2 θ sin 2 θ + 2 sin θ cos θ

= a 2 cos 2θ + sin 2θ
= 2a 2 2 2 cos 2θ + sin 2θ 2 2

π = 2a 2 sin 2θ + 4
≤ 2a 2 ≤ 2

x 2 + y 2 ≤ 1 的情况下设 x = a cos θ , y = a sin θ

(0 ≤ a ≤ 1) 是常用的换

b c 1 + ≥ 2 c+d a+b 2

b c + 形式不变, c+d a+b 只是 a+b 和 c+d 彼此换位,所以不失一般性可以认为 a + b ≥ c + d

b c b+c 1 1 + = c c+d a+b a+d c + d a +b

b+c≥ b c + c+d a+b ≥ =

1 (a + b + c + d ) , 2

1 a+b+c+d 1 1 (c + d ) 2 c+d c + d a +b 1 a+b c+d 1 + 2 c+d a+b 2

≥2

1 a+b c+d 1 1 = 2 2 c+d a+b 2 2

1 是可达到的,如 a = 2 + 1, b = 2 1, c = 2, d = 0 即可 2

2 3 4 n < 3

k

(k + 1)

n 1 n
ka k +1 ,则有

a k +1

2 a k k 2 + 2k + 1 = ≥ >k +2, k k

(n 1)

n ≥n

2 3 4 n < 3

2 2 a12 + a 2 + + a n a1 + a 2 + + a n a2 ≤ , + 4 n n 2

a1 + a 2 + + a n = m ,则 n
2

2 2 a12 + a 2 + + a n a2 a a2 m 2 ≤ am m 2 = m ≤ 4 2 4 n

2 2 2 a12 + a 2 + + a n a1 + a 2 + + a n a2 所以 ≤ + n n 4 2

2

a ,而这只在下列两种情况下才可能:①所有的 a i 都为 0,②m 2 n n 为偶数, 个 a i = 0 ,其余的 个 a i = a (a > 0 ) 2 2

a i = a .第二, m =

4 4 4 2 2 2 2 2 2 若(1) a + b + c ≤ 2 a b + b c + c a 成立

(

)

2 2 2

b) 反过来,不等式(2)成立能推出(1)也成立吗? 证:a)设

D = 2 a 2b 2 + b 2 c 2 + c 2 a 2 a 4 + b 4 + c 4 = 4a 2 b 2 a 2 + b 2
2

(

) ( )
2

)

(

)

2

+ 2c 2 a 2 + b 2 c 4

(

)

= (2ab ) a 2 + b 2 c 2

(

a 2 + b 2 + c 2 = a 2 + b 2 c 2 + 2c 2 ≤ 2ab + 2c 2 ≤ 2(ab + bc + ca )
b) 例如 a = 4, b = c = 1 时,

a 2 + b 2 + c 2 = 18 ≤ 2(4 1 + 1 1 + 1 4 ) = 18 ,
(2)成立.但

a 4 + b 4 + c 4 = 258 ≥ 66 = 2 4 2 12 + 12 12 + 12 4 2 ,

(

)

3 ≤ x + y + y +1 + 2y x 4 ≤ 7

1 ≤ y ≤ 1 , ∴ y + 1 ≥ ( 1) + 1 = 0

y +1 = y +1
∵ x 2 ≤ 1 1 ≤ x ≤ 1

2 × ( 1) 1 ≤ 2 y x ≤ 2 × 1 ( 1) ,
3 ≤ 2y x ≤ 3,

2 y x 4 ≤ 3 4 = 1 < 0 ,

2 y x 4 = (2 y x 4) = 2 y + x + 4

x + y + y +1 + 2y x 4

= ( x + y ) + ( y + 1) (2 y x 4) = 2 x + 5 ∴ 3 = 2 × ( 1) + 5 ≤ 2 x + 5 ≤ 2 × 1 + 5 = 7
(2) x + y < 0 时,

x + y + y +1 + 2y x 4 = ( x + y ) + ( y + 1) (2 y x 4) = 2 y + 5 , ∴ 3 = 2 × 1 + 5 ≤ 2 y + 5 ≤ 2 × ( 1) + 5 = 7

3 ≤ x + y + y +1 + 2y x 4 ≤ 7

π
2

,0 < y <

π
2

,0 < θ <

π
2

, θ < x ,且

2 sin 2 x + sin 2 y = 3 sin 2 θ = 1 ,

π
2

,0 < x <

π
4

.

2 sin 2 θ + sin 2 y < 2 sin 2 x + sin 2 y = 3 sin 2 θ = 2 sin 2 θ + sin 2 θ
∴ sin 2 y < sin 2 θ y < θ

0 < y <θ < x <

π
4

(*)

2 sin 2 x + sin 2 y = 3 sin 2 θ , 2 sin 2 x sin 2 θ = sin 2 θ sin 2 y ,

(

)

2 sin ( x + θ ) sin ( x θ ) = sin (θ + y ) sin (θ y )
2 sin ( x θ ) sin (θ + y ) = , sin (θ y ) sin ( x + θ )

2 sin ( x θ ) cos( x θ ) sin (θ + y ) cos( x θ ) = sin (θ y ) sin ( x + θ )

sin 2( x θ ) sin (θ + y ) cos( x θ ) = <1 sin (θ y ) sin ( x + θ )

2 x + y < 3θ

7 27 3 z ≤ x + y + z = 1 ,即 z ≤ 1 3

2 xyz ≤

2 xy ≤ xy , 3

0 ≤ xy + yz + zx 2 xyz

2y ≤ x + y ≤ x + y + z =1

y≤

1 2 1 ,从而 3

xy + yz + zx 2 xyz = y ( z + x ) + zx(1 2 y )
1 1 ≤ y ( z + x ) + zx(1 2 y ) + x z (1 2 y ) 3 3

1 1 = y (z + x ) + x + z (1 2 y ) 3 3 1 1 = y ω + + ω (1 2 y ) 3 3
=
(其中 ω = x + z

1 ) 3

1 1 yω + ( y + ω ) 3 3 1 2 y +ω = y + x + z = , 3 3 1 1 2 yω ≤ ( y + ω ) = , 4 9 1 1 1 2 7 xy + yz + xz 2 xyz ≤ + = 3 9 3 3 27

2 2 2 xn1 xn x12 x2 + + + + ≥ x1 + x2 + + xn x2 x3 xn x1

x2 x2 x12 x2 + x 2 ≥ 2 x1 , 2 + x3 ≥ 2 x 2 , , n 1 + x n ≥ 2 x n 1 , n + x1 ≥ 2 x n x2 x3 xn x1

2 2 2 x12 x 2 x n 1 x n x + x + + x + x + ( x1 + x 2 + + x n ) ≥ 2( x1 + x 2 + + x n ) 3 n 1 2 2 x2 x2 x12 x 2 + + + n 1 + n ≥ x1 + x 2 + + x n 1 + x n x 2 x3 xn x1
2
2 2

m(m n ) ≥ n(m n )
n ≥ 0, m ∴ (m n ) ≥ m n n

(*)

x1 x (x1 x 2 ) ≥ x1 x 2 , 2 (x 2 x3 ) ≥ x 2 x3 , , x2 x3 x n 1 x (x n1 x n ) ≥ x n 1 x n , n (x n x1 ) ≥ x n x1 , xn x1

x1 x x x (x1 x 2 ) + 2 (x 2 x3 ) + + n1 (x n1 x n ) + n (x n x1 ) x2 x3 xn x1
≥ ( x1 x 2 ) + ( x 2 x3 ) + + ( x n 1 x n ) + ( x n x1 ) = 0

2 2 2 x n 1 x n x12 x 2 + ++ + ≥ x1 + x 2 + + x n 1 + x n x2 x3 xn x1

2 2 2 2 x n x12 x12 x 2 x2 x 2 x 2 + x12 x12 + x 2 x 2 + xn + + + + + n 1 + n = n + + + n 1 x1 x1 x 2 x 2 xn xn x1 x2 xn

x x x x xx ≥ 2 1 n + 1 2 + + n 1 n x x2 xn 1
= 2( x1 + x 2 + + x n )

2 2 2 x n 1 x n x12 x 2 + ++ + ≥ x1 + x 2 + + x n 1 + x n 成立 x2 x3 xn x1

a1 = x 2 , a 2 = x3 , a 3 = x 4 , , a n 1 = x n , a n = x1 ,
b1 = x1 x2 , b2 = x2 x3 , b3 = x3 x4 , , bn 1 = x n 1 xn
2

, bn =

xn x1

n 2 n 2 n 根据柯西不等式 ∑ a i ∑ bi ≥ ∑ a i bi ,得 i =1 i =1 i =1
x 2

( ) +( x )
2 3

2

++

( x ) +( x )
2 2 n 1

2 2 2 2 x2 x n 1 xn x1 ++ + + x x x x 2 1 3 n

x x x x ≥ x 2 1 + x3 2 + + x n n 1 + x1 n , x2 x3 xn x1

2

(x 2 + x3 + + x n + x1 )

x12 x2
2

+

2 x2 x2 x2 + + n 1 + n x3 xn x1

≥ ( x1 + x 2 + + x n 1 + x n )

∵ x1 + x 2 + + x n 1 + x n > 0

2 x2 x2 x12 x 2 + + + n 1 + n ≥ x1 + x 2 + + x n 1 + x n x2 x3 xn x1

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