Example 2.1 It is known that a > b > c a > b > c a > b > c
1 a − b + 1 b − c ≥ k a − c \frac{1}{a-b} + \frac{1}{b-c} \geq \frac{k}{a-c} a − b 1 + b − c 1 ≥ a − c k
The maximum value of a constant real number k k k
The denominator is complicated, so change the denominator:
a − b = x , b − c = y , a − c = x + y 1 x + 1 y ≥ ( 1 + 1 ) 2 x + y ⟹ k ≤ 4 \begin{gathered} a-b=x,b-c=y,a-c=x+y\\ \frac{1}{x}+\frac{1}{y}\ge\frac{(1+1)^2}{x+y}\Longrightarrow k\le4 \end{gathered} a − b = x , b − c = y , a − c = x + y x 1 + y 1 ≥ x + y ( 1 + 1 ) 2 ⟹ k ≤ 4
Example 2.2 Assume a , b , c a, b, c a , b , c
a + 3 c a + 2 b + c + 4 b a + b + 2 c − 8 c a + b + 3 c ≥ 12 2 − 17. \frac{a + 3c}{a + 2b + c} + \frac{4b}{a + b + 2c} - \frac{8c}{a + b + 3c} \geq 12\sqrt{2} - 17. a + 2 b + c a + 3 c + a + b + 2 c 4 b − a + b + 3 c 8 c ≥ 12 2 − 17.
The denominator is complicated, so change the denominator:
a + 2 b + c = x , a + b + 2 c = y , a + b + 3 c = z { a = − x + 5 y − 3 z , b = x − 2 y + z , c = 0 x − y + z − x + 2 y x + 4 x − 8 y + 4 z y + 8 y − 8 z z = − 17 + 2 y x + 4 x y + 4 z y + 8 y z ≥ − 17 + 2 2 ⋅ 4 + 2 4 ⋅ 8 = − 17 + 12 2 \begin{gathered} a+2b+c=x,a+b+2c=y,a+b+3c=z\\ \begin{cases} a=-x+5y-3z,\\ b=x-2y+z,\\ c=0x-y+z \end{cases}\\ \frac{-x+2y}{x}+\frac{4x-8y+4z}{y}+\frac{8y-8z}{z}\\ =-17+2\frac{y}{x}+4\frac{x}{y}+4\frac{z}{y}+8\frac{y}{z}\\ \ge-17+2\sqrt{2\cdot4}+2\sqrt{4\cdot8}=-17+12\sqrt{2} \end{gathered} a + 2 b + c = x , a + b + 2 c = y , a + b + 3 c = z ⎩ ⎨ ⎧ a = − x + 5 y − 3 z , b = x − 2 y + z , c = 0 x − y + z x − x + 2 y + y 4 x − 8 y + 4 z + z 8 y − 8 z = − 17 + 2 x y + 4 y x + 4 y z + 8 z y ≥ − 17 + 2 2 ⋅ 4 + 2 4 ⋅ 8 = − 17 + 12 2
Example 2.3 Given a > b > 0 a > b > 0 a > b > 0 a 2 + 1 b ( a − b ) a^2 + \frac{1}{b(a-b)} a 2 + b ( a − b ) 1
A. 2
B. 4
C. 2 5 2\sqrt{5} 2 5
D. 5
Consider b first: a 2 + 1 b ( a − b ) ≥ a 2 + 4 a 2 ≥ 4 \begin{gathered} a^2 + \frac{1}{b(a-b)}\\ \ge a^2+\frac{4}{a^2}\ge4 \end{gathered} a 2 + b ( a − b ) 1 ≥ a 2 + a 2 4 ≥ 4
Or consider incremental substitution :
b = x , a = x + y ( x + y ) 2 + 1 x y ≥ 4 x y + 1 x y ≥ 4 \begin{gathered} b=x,a=x+y\\ (x+y)^2+\frac{1}{xy}\\ \ge4xy+\frac{1}{xy}\ge4 \end{gathered} b = x , a = x + y ( x + y ) 2 + x y 1 ≥ 4 x y + x y 1 ≥ 4
Example 2.4 (2012 Tsinghua Summer Camp) It is known that a , b , c a, b, c a , b , c △ A B C \triangle ABC △ A B C
A. 3 2 ≤ a b + c + b c + a + c a + b < 2 \frac{3}{2} \leq \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} < 2 2 3 ≤ b + c a + c + a b + a + b c < 2
B. 3 2 < a b + c + b c + a + c a + b < 2 \frac{3}{2} \lt \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} < 2 2 3 < b + c a + c + a b + a + b c < 2
C. 3 2 < a b + c + b c + a + c a + b ≤ 2 \frac{3}{2} \lt \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \leq 2 2 3 < b + c a + c + a b + a + b c ≤ 2
D. 3 2 ≤ a b + c + b c + a + c a + b ≤ 2 \frac{3}{2} \leq \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \leq 2 2 3 ≤ b + c a + c + a b + a + b c ≤ 2
a < b + c ⟹ 2 a a + b + c > a b + c \begin{gathered} a\lt b+c\Longrightarrow \frac{2a}{a+b+c}\gt\frac{a}{b+c}\\ \end{gathered} a < b + c ⟹ a + b + c 2 a > b + c a
Rotate a → b → c → a a \to b \to c \to a a → b → c → a
b < c + a ⟹ 2 b a + b + c > b c + a b < c+a \quad\Longrightarrow\quad \frac{2b}{a+b+c} > \frac{b}{c+a} b < c + a ⟹ a + b + c 2 b > c + a b
c < a + b ⟹ 2 c a + b + c > c a + b c < a+b \quad\Longrightarrow\quad \frac{2c}{a+b+c} > \frac{c}{a+b} c < a + b ⟹ a + b + c 2 c > a + b c
Accumulation: S < 2 S\lt2 S < 2
Let's consider the lower bound, repeat the same trick, change the denominator, and use the AM-GM inequality:
b + c = x , c + a = y , a + b = z S = y + z − x 2 x + z + x − y 2 y + x + y − z 2 z = 1 2 ( x y + y x + y z + z y + z x + x z ) − 3 2 ≥ 3 − 3 2 = 3 2 \begin{gathered} b+c=x,c+a=y,a+b=z\\ S=\frac{\frac{y+z-x}{2}}{x}+\frac{\frac{z+x-y}{2}}{y}+\frac{\frac{x+y-z}{2}}{z}\\ =\frac{1}{2}(\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{z}{x}+\frac{x}{z})-\frac{3}{2}\\ \ge 3-\frac{3}{2}=\frac{3}{2} \end{gathered} b + c = x , c + a = y , a + b = z S = x 2 y + z − x + y 2 z + x − y + z 2 x + y − z = 2 1 ( y x + x y + z y + y z + x z + z x ) − 2 3 ≥ 3 − 2 3 = 2 3
Choose A.
Example 2.5 If the positive number x , y x, y x , y x 2 + 2 x y − 1 = 0 x^2 + 2xy - 1 = 0 x 2 + 2 x y − 1 = 0 2 x + y 2x + y 2 x + y
A. 2 2 \frac{\sqrt{2}}{2} 2 2
B. 2 \sqrt{2} 2
C. 3 2 \frac{\sqrt{3}}{2} 2 3
D. 3 \sqrt{3} 3
2 x + y = u , y = u − 2 x x 2 + 2 ( u − 2 x ) x − 1 = 0 3 x 2 − 2 u x + 1 = 0 Δ = 4 u 2 − 12 ≥ 0 u ≥ 3 ( x = y = 3 3 ) \begin{gathered} 2x+y=u,y=u-2x\\ x^2+2(u-2x)x-1=0\\ 3x^2-2ux+1=0\\ \Delta=4u^2-12\ge0\\ u\ge\sqrt{3}(x=y=\frac{\sqrt{3}}{3}) \end{gathered} 2 x + y = u , y = u − 2 x x 2 + 2 ( u − 2 x ) x − 1 = 0 3 x 2 − 2 ux + 1 = 0 Δ = 4 u 2 − 12 ≥ 0 u ≥ 3 ( x = y = 3 3 )
Or consider exchanging yuan: x ( x + 2 y ) = 1 2 x + y = 3 2 x + 1 2 ( x + 2 y ) ≥ 3 x ( x + 2 y ) = 3 \begin{gathered} x(x+2y)=1\\ 2x+y=\frac{3}{2}x+\frac{1}{2}(x+2y)\ge\sqrt{3x(x+2y)}=\sqrt{3} \end{gathered} x ( x + 2 y ) = 1 2 x + y = 2 3 x + 2 1 ( x + 2 y ) ≥ 3 x ( x + 2 y ) = 3
Example 2.6 If the positive number a , b , c a, b, c a , b , c a ( a + b + c ) + b c = 4 a(a+b+c)+bc=4 a ( a + b + c ) + b c = 4 3 a + 2 b + c 3a+2b+c 3 a + 2 b + c
a 2 + a b + a c + b c = ( a + b ) ( a + c ) = 4 3 a + 2 b + c = 2 ( a + b ) + ( a + c ) ≥ 2 2 ( a + b ) ( a + c ) = 4 3 \begin{gathered} a^2+ab+ac+bc=(a+b)(a+c)=4\\ 3a+2b+c=2(a+b)+(a+c)\\\ge2\sqrt{2(a+b)(a+c)}=4\sqrt{3} \end{gathered} a 2 + ab + a c + b c = ( a + b ) ( a + c ) = 4 3 a + 2 b + c = 2 ( a + b ) + ( a + c ) ≥ 2 2 ( a + b ) ( a + c ) = 4 3
Example 2.7 Assume x , y , z > 0 x, y, z > 0 x , y , z > 0
x y + 2 y z x 2 + y 2 + z 2 \frac{xy + 2yz}{x^2 + y^2 + z^2} x 2 + y 2 + z 2 x y + 2 y z
The maximum value is ______.
x y + 2 y z x 2 + y 2 + z 2 = x y + 2 y z x 2 + 1 5 y 2 + 4 5 y 2 + z 2 ≤ x y + 2 y z 2 1 5 x y + 2 4 5 y z ≤ 5 2 \begin{gathered} \frac{xy + 2yz}{x^2 + y^2 + z^2}\\ =\frac{xy + 2yz}{x^2 + \frac{1}{5}y^2+\frac{4}{5}y^2 + z^2}\\ \le \frac{xy+2yz}{2\sqrt{\frac{1}{5}}xy+2\sqrt{\frac{4}{5}yz}}\\ \le\frac{\sqrt{5}}{2} \end{gathered} x 2 + y 2 + z 2 x y + 2 y z = x 2 + 5 1 y 2 + 5 4 y 2 + z 2 x y + 2 y z ≤ 2 5 1 x y + 2 5 4 y z x y + 2 y z ≤ 2 5
Example 2.8 Given a ≥ 0 , b ≥ 0 , c ≥ 0 , a + b + c = 1 a \geq 0, b \geq 0, c \geq 0, a + b + c = 1 a ≥ 0 , b ≥ 0 , c ≥ 0 , a + b + c = 1 4 a + 1 + 4 b + 1 + 4 c + 1 \sqrt{4a+1} + \sqrt{4b+1} + \sqrt{4c+1} 4 a + 1 + 4 b + 1 + 4 c + 1
Round up the mean by averaging:
( 4 a + 1 ) 7 3 ≤ 4 a + 1 + 7 3 2 \begin{gathered} \sqrt{(4a+1)\frac{7}{3}}\le\frac{4a+1+\frac{7}{3}}{2} \end{gathered} ( 4 a + 1 ) 3 7 ≤ 2 4 a + 1 + 3 7 4 a + 1 ≤ 2 21 7 a + 5 21 7 4 b + 1 ≤ 2 21 7 b + 5 21 7 4 c + 1 ≤ 2 21 7 c + 5 21 7 S ≤ 2 21 7 ( a + b + c ) + 5 21 7 = 21 \begin{gathered} \sqrt{4a+1}\le\frac{2\sqrt{21}}{7}a+\frac{5\sqrt{21}}{7}\\ \sqrt{4b+1}\le\frac{2\sqrt{21}}{7}b+\frac{5\sqrt{21}}{7}\\ \sqrt{4c+1}\le\frac{2\sqrt{21}}{7}c+\frac{5\sqrt{21}}{7}\\ S\le\frac{2\sqrt{21}}{7}(a+b+c)+\frac{5\sqrt{21}}{7}=\sqrt{21} \end{gathered} 4 a + 1 ≤ 7 2 21 a + 7 5 21 4 b + 1 ≤ 7 2 21 b + 7 5 21 4 c + 1 ≤ 7 2 21 c + 7 5 21 S ≤ 7 2 21 ( a + b + c ) + 7 5 21 = 21
Example 2.9 (Zhejiang University) Suppose the sum of positive numbers x 1 , x 2 , … , x n x_1, x_2, \dots, x_n x 1 , x 2 , … , x n n ≥ 2 n \geq 2 n ≥ 2
1 x 1 − x 1 3 + 1 x 2 − x 2 3 + ⋯ + 1 x n − x n 3 > 4 \frac{1}{x_1 - x_1^3} + \frac{1}{x_2 - x_2^3} + \cdots + \frac{1}{x_n - x_n^3} > 4 x 1 − x 1 3 1 + x 2 − x 2 3 1 + ⋯ + x n − x n 3 1 > 4
Consider local inequalities:
1 x − x 3 ≥ 4 x , x ∈ ( 0 , 1 ) ⟺ ( 2 x 2 − 1 ) 2 ≥ 0 ( x = 2 2 ) \begin{gathered} \frac{1}{x-x^3}\ge4x,x\in(0,1)\\ \Longleftrightarrow (2x^2-1)^2\ge0(x=\frac{\sqrt{2}}{2}) \end{gathered} x − x 3 1 ≥ 4 x , x ∈ ( 0 , 1 ) ⟺ ( 2 x 2 − 1 ) 2 ≥ 0 ( x = 2 2 )
Obviously, it is impossible for every x x x 2 2 \frac{\sqrt{2}}{2} 2 2
Example 2.10 Given a i > 0 a_i > 0 a i > 0 a 1 a 2 a 3 ⋯ a n = 1 a_1a_2a_3\cdots a_n = 1 a 1 a 2 a 3 ⋯ a n = 1 ( a 1 + 2 ) ( a 2 + 2 ) ⋯ ( a n + 2 ) (a_1+2)(a_2+2)\cdots(a_n+2) ( a 1 + 2 ) ( a 2 + 2 ) ⋯ ( a n + 2 )
Still holding the hand holding the equal sign: ( a 1 + 1 + 1 ) ( a 2 + 1 + 1 ) ⋯ ( a n + 1 + 1 ) ≥ 3 n a 1 a 2 a 3 ⋯ a n 3 = 3 n \begin{gathered} (a_1+1+1)(a_2+1+1)\cdots(a_n+1+1)\\ \ge3^n\sqrt[3]{a_1a_2a_3\cdots a_n}=3^n \end{gathered} ( a 1 + 1 + 1 ) ( a 2 + 1 + 1 ) ⋯ ( a n + 1 + 1 ) ≥ 3 n 3 a 1 a 2 a 3 ⋯ a n = 3 n
Example 2.11 Given x > 0 x > 0 x > 0 x 2 + 16 x x^2 + \frac{16}{x} x 2 + x 16
x 2 + 8 x + 8 x ≥ 3 8 2 3 = 12 ( x = 2 ) \begin{gathered} x^2+\frac{8}{x}+\frac{8}{x}\\ \ge3\sqrt[3]{8^2}=12(x=2) \end{gathered} x 2 + x 8 + x 8 ≥ 3 3 8 2 = 12 ( x = 2 )
Example 2.12 Given x > 0 x\gt0 x > 0 x 2 + 1 x 3 x^2+\frac{1}{x^3} x 2 + x 3 1
x 2 3 + x 2 3 + x 2 3 + 1 2 x 3 + 1 2 x 3 ≥ 5 1 3 3 ⋅ 2 2 5 = 5 6 72 5 \begin{gathered} \frac{x^2}{3}+\frac{x^2}{3}+\frac{x^2}{3}+\frac{1}{2x^3}+\frac{1}{2x^3}\\ \ge5\sqrt[5]{\frac{1}{3^3\cdot2^2}}=\frac{5}{6}\sqrt[5]{72} \end{gathered} 3 x 2 + 3 x 2 + 3 x 2 + 2 x 3 1 + 2 x 3 1 ≥ 5 5 3 3 ⋅ 2 2 1 = 6 5 5 72
Example 2.13 Given 0 < x < 1 0\lt x\lt1 0 < x < 1 x 2 1 − x x^2\sqrt{1-x} x 2 1 − x
The lower bound is obviously 0, there is no minimum value, consider the maximum value:
x 2 1 − x = x 4 ( 1 − x ) = 1 2 x ⋅ x ⋅ x ⋅ x ( 4 − 4 x ) ≤ 1 2 ( 4 5 ) 5 = 16 5 125 ( x = 4 5 ) \begin{gathered} x^2\sqrt{1-x}\\ =\sqrt{x^4(1-x)}\\ =\frac{1}{2}\sqrt{x\cdot x\cdot x\cdot x(4-4x)}\le\frac{1}{2}\sqrt{(\frac{4}{5})^5}=\frac{16\sqrt{5}}{125}(x=\frac{4}{5}) \end{gathered} x 2 1 − x = x 4 ( 1 − x ) = 2 1 x ⋅ x ⋅ x ⋅ x ( 4 − 4 x ) ≤ 2 1 ( 5 4 ) 5 = 125 16 5 ( x = 5 4 )
Example 2.14 Assume x > − 1 2 x > -\frac{1}{2} x > − 2 1 f ( x ) = x 2 + x + 4 2 x + 1 f(x) = x^2 + x + \frac{4}{2x+1} f ( x ) = x 2 + x + 2 x + 1 4
Change denominator: t = 2 x + 1 > 0 t=2x+1\gt0 t = 2 x + 1 > 0
f ( x ) = ( t − 1 2 ) 2 + t − 1 2 + 4 t = t 2 4 + 4 t − 1 4 = t 2 4 + 2 t + 2 t − 1 4 ≥ 3 − 1 4 = 11 4 ( t = 2 , x = 1 2 ) \begin{gathered} f(x)=(\frac{t-1}{2})^2+\frac{t-1}{2}+\frac{4}{t}\\ =\frac{t^2}{4}+\frac{4}{t}-\frac{1}{4}\\ =\frac{t^2}{4}+\frac{2}{t}+\frac{2}{t}-\frac{1}{4}\ge3-\frac{1}{4}=\frac{11}{4}(t=2,x=\frac{1}{2}) \end{gathered} f ( x ) = ( 2 t − 1 ) 2 + 2 t − 1 + t 4 = 4 t 2 + t 4 − 4 1 = 4 t 2 + t 2 + t 2 − 4 1 ≥ 3 − 4 1 = 4 11 ( t = 2 , x = 2 1 )
Example 2.15 Assume a + 3 b = 3 a + 3b = 3 a + 3 b = 3 3 a + 9 b 3^a + 9^b 3 a + 9 b
3 a + 9 b = 3 a + 3 2 b = 3 a 2 + 3 a 2 + 3 2 b 3 + 3 2 b 3 + 3 2 b 3 ≥ 5 3 2 a + 6 b 2 2 ⋅ 3 3 5 = 5 3 3 2 2 5 = 5 2 ( 15 ) 3 5 \begin{gathered} 3^a+9^b=3^a+3^{2b}\\ =\frac{3^a}{2}+\frac{3^a}{2}+\frac{3^{2b}}{3}+\frac{3^{2b}}{3}+\frac{3^{2b}}{3}\\ \ge5\sqrt[5]{\frac{3^{2a+6b}}{2^2\cdot3^3}}=5\sqrt[5]{\frac{3^3}{2^2}}=\frac{5}{2}(15)^\frac{3}{5} \end{gathered} 3 a + 9 b = 3 a + 3 2 b = 2 3 a + 2 3 a + 3 3 2 b + 3 3 2 b + 3 3 2 b ≥ 5 5 2 2 ⋅ 3 3 3 2 a + 6 b = 5 5 2 2 3 3 = 2 5 ( 15 ) 5 3
Example 2.16 Assume a > b > 0 a > b > 0 a > b > 0 2 a 3 + 3 a b − b 2 ≥ 10 \sqrt{2a^3} + \frac{3}{ab - b^2} \ge 10 2 a 3 + ab − b 2 3 ≥ 10
① ≥ 2 ⋅ ( 2 x y ) 3 + 3 x y \ge \sqrt{2} \cdot (2\sqrt{xy})^3 + \frac{3}{xy} ≥ 2 ⋅ ( 2 x y ) 3 + x y 3 = 8 2 ( x y ) 3 2 + 3 x y = 8\sqrt{2}(xy)^{\frac{3}{2}} + \frac{3}{xy} = 8 2 ( x y ) 2 3 + x y 3 = 4 2 ( x y ) 3 2 + 4 2 ( x y ) 3 2 + 3 x y = 4\sqrt{2}(xy)^{\frac{3}{2}} + 4\sqrt{2}(xy)^{\frac{3}{2}} + \frac{3}{xy} = 4 2 ( x y ) 2 3 + 4 2 ( x y ) 2 3 + x y 3 ≥ 5 32 5 = 10 \ge 5\sqrt[5]{32} = 10 ≥ 5 5 32 = 10 ≥ 2 ( x + y ) 3 + 12 ( x + y ) 2 \ge \sqrt{2}(x+y)^3 + \frac{12}{(x+y)^2} ≥ 2 ( x + y ) 3 + ( x + y ) 2 12 = 2 2 ( x + y ) 3 + 2 2 ( x + y ) 3 + 4 ( x + y ) 2 + 4 ( x + y ) 2 + 4 ( x + y ) 2 = \frac{\sqrt{2}}{2}(x+y)^3 + \frac{\sqrt{2}}{2}(x+y)^3 + \frac{4}{(x+y)^2} + \frac{4}{(x+y)^2} + \frac{4}{(x+y)^2} = 2 2 ( x + y ) 3 + 2 2 ( x + y ) 3 + ( x + y ) 2 4 + ( x + y ) 2 4 + ( x + y ) 2 4 ≥ 5 32 5 = 10 \ge 5\sqrt[5]{32} = 10 ≥ 5 5 32 = 10
Example 2.17 Assume a , b , c a,b,c a , b , c a + b + c = 1 a + b + c = 1 a + b + c = 1 1 a + b c + 1 b + a c + 1 c + a b ≥ 27 4 \frac{1}{a+bc} + \frac{1}{b+ac} + \frac{1}{c+ab} \ge \frac{27}{4} a + b c 1 + b + a c 1 + c + ab 1 ≥ 4 27
The equal sign is obviously a = b = c = 1 3 a=b=c=\frac{1}{3} a = b = c = 3 1 1 a + b c = 1 a ( a + b + c ) + b c = 1 ( a + b ) ( a + c ) S ≥ 3 1 3 ( a + b ) ( a + c ) ( b + c ) ( b + a ) ( c + a ) ( c + b ) 3 ≥ 3 1 ( 4 6 ) 6 3 = 27 4 \begin{gathered} \frac{1}{a+bc}\\ =\frac{1}{a(a+b+c)+bc}\\ =\frac{1}{(a+b)(a+c)}\\ S\ge3\sqrt[3]{\frac{1^3}{(a+b)(a+c)(b+c)(b+a)(c+a)(c+b)}}\\ \ge3\sqrt[3]{\frac{1}{(\frac{4}{6})^6}}=\frac{27}{4} \end{gathered} a + b c 1 = a ( a + b + c ) + b c 1 = ( a + b ) ( a + c ) 1 S ≥ 3 3 ( a + b ) ( a + c ) ( b + c ) ( b + a ) ( c + a ) ( c + b ) 1 3 ≥ 3 3 ( 6 4 ) 6 1 = 4 27
Example 2.18 It is known that the positive real number a , b a,b a , b a b ( a + b ) = 4 ab(a+b) = 4 ab ( a + b ) = 4 2 a + b 2a + b 2 a + b ‾ \underline{\quad\quad}
4 u v = ( u a ) ( v b ) ( a + b ) ≤ ( ( u + 1 ) a + ( v + 1 ) b 3 ) 3 u + 1 v + 1 = 2 , u a = v b = a + b = k u = 2 v + 1 , a = k 2 v + 1 , b = k v 1 v + 1 2 v + 1 = 1 ⟹ v = 1 + 3 2 , u = 2 + 3 3 + 3 2 ( 2 a + b ) 3 ≥ 2 ( 3 + 1 ) ( 3 + 2 ) 3 = 1 + 3 2 a + b ≥ 2 3 \begin{gathered} 4uv=(ua)(vb)(a+b)\le(\frac{(u+1)a+(v+1)b}{3})^3\\ \frac{u+1}{v+1}=2,ua=vb=a+b=k\\ u=2v+1,a=\frac{k}{2v+1},b=\frac{k}{v}\\ \frac{1}{v}+\frac{1}{2v+1}=1\\ \Longrightarrow v=\frac{1+\sqrt{3}}{2},u=2+\sqrt{3}\\ \frac{\frac{3+\sqrt{3}}{2}(2a+b)}{3}\ge \sqrt[3]{2(\sqrt{3}+1)(\sqrt{3}+2)}=1+\sqrt{3}\\ 2a+b\ge2\sqrt{3} \end{gathered} 4 uv = ( u a ) ( v b ) ( a + b ) ≤ ( 3 ( u + 1 ) a + ( v + 1 ) b ) 3 v + 1 u + 1 = 2 , u a = v b = a + b = k u = 2 v + 1 , a = 2 v + 1 k , b = v k v 1 + 2 v + 1 1 = 1 ⟹ v = 2 1 + 3 , u = 2 + 3 3 2 3 + 3 ( 2 a + b ) ≥ 3 2 ( 3 + 1 ) ( 3 + 2 ) = 1 + 3 2 a + b ≥ 2 3
Or a more attention-demanding solution:
( 2 a + b ) 2 = 4 a ( a + b ) + b 2 ≥ 16 b + b 2 ≥ 12 ⟹ 2 a + b ≥ 2 3 \begin{gathered} (2a+b)^2=4a(a+b)+b^2\ge\frac{16}{b}+b^2\ge12\\ \Longrightarrow 2a+b\ge2\sqrt{3} \end{gathered} ( 2 a + b ) 2 = 4 a ( a + b ) + b 2 ≥ b 16 + b 2 ≥ 12 ⟹ 2 a + b ≥ 2 3
Example 2.19 Assume S n = ∑ k = 1 n 1 k S_n = \sum_{k=1}^n \frac{1}{k} S n = ∑ k = 1 n k 1 n ( n + 1 ) 1 / n − n < S n < n − ( n − 1 ) n − 1 / ( n − 1 ) ( n > 1 ) n(n+1)^{1/n} - n < S_n < n - (n-1)n^{-1/(n-1)} \quad (n > 1) n ( n + 1 ) 1/ n − n < S n < n − ( n − 1 ) n − 1/ ( n − 1 ) ( n > 1 )
Key:1 pairing S n + n = 2 1 + 3 2 + 4 3 + ⋯ + n + 1 n ≥ n ( n + 1 ) 1 n n − S n = 0 + 1 2 + 2 3 + ⋯ + n − 1 n ≥ ( n − 1 ) ( 1 n ) 1 n − 1 \begin{gathered} S_n+n=\frac{2}{1}+\frac{3}{2}+\frac{4}{3}+\cdots+\frac{n+1}{n}\ge n(n+1)^\frac{1}{n}\\ n-S_n=0+\frac{1}{2}+\frac{2}{3}+\cdots+\frac{n-1}{n}\ge(n-1)(\frac{1}{n})^\frac{1}{n-1} \end{gathered} S n + n = 1 2 + 2 3 + 3 4 + ⋯ + n n + 1 ≥ n ( n + 1 ) n 1 n − S n = 0 + 2 1 + 3 2 + ⋯ + n n − 1 ≥ ( n − 1 ) ( n 1 ) n − 1 1
Example 2.20 Proof: 1 − 1 2012 ( 1 2 + 1 3 + ⋯ + 1 2013 ) > 1 2013 2012 1 - \frac{1}{2012} \left( \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2013} \right) > \frac{1}{\sqrt[2012]{2013}} 1 − 2012 1 ( 2 1 + 3 1 + ⋯ + 2013 1 ) > 2012 2013 1
Still note the pairing of 1: ⟺ 2012 − ( 1 2 + 1 3 + ⋯ + 1 2013 ) > 2012 1 2023 2013 ⟺ 1 2 + 2 3 + 3 4 + ⋯ + 2012 2013 > 2012 1 2023 2013 \begin{gathered} \Longleftrightarrow 2012-(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2013})\gt 2012\sqrt[2013]{\frac{1}{2023}}\\ \Longleftrightarrow \frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\cdots+\frac{2012}{2013}\gt 2012\sqrt[2013]{\frac{1}{2023}} \end{gathered} ⟺ 2012 − ( 2 1 + 3 1 + ⋯ + 2013 1 ) > 2012 2013 2023 1 ⟺ 2 1 + 3 2 + 4 3 + ⋯ + 2013 2012 > 2012 2013 2023 1
Example 2.21 It is known that m , n m, n m , n 1 < m < n 1 < m < n 1 < m < n ( 1 + m ) n > ( 1 + n ) m (1+m)^n > (1+n)^m ( 1 + m ) n > ( 1 + n ) m ( 1 + n ) m ⋅ 1 × 1 × ⋯ × 1 ⏟ n − m terms < ( ( n − m ) + m ( 1 + n ) n ) n = ( 1 + m ) n \begin{gathered} (1+n)^m\cdot\underbrace{1 \times 1 \times \cdots \times 1}_{n-m\text{ terms}}\\ \lt(\frac{(n-m)+m(1+n)}{n})^n=(1+m)^n \end{gathered} ( 1 + n ) m ⋅ n − m terms 1 × 1 × ⋯ × 1 < ( n ( n − m ) + m ( 1 + n ) ) n = ( 1 + m ) n
Example 2.22 Suppose x 1 , x 2 , … , x n + 1 > 0 x_1, x_2, \dots, x_{n+1} > 0 x 1 , x 2 , … , x n + 1 > 0 1 1 + x 1 + 1 1 + x 2 + ⋯ + 1 1 + x n + 1 = 1 \frac{1}{1+x_1} + \frac{1}{1+x_2} + \dots + \frac{1}{1+x_{n+1}} = 1 1 + x 1 1 + 1 + x 2 1 + ⋯ + 1 + x n + 1 1 = 1 x 1 x 2 ⋯ x n + 1 ≥ n n + 1 x_1x_2\cdots x_{n+1} \ge n^{n+1} x 1 x 2 ⋯ x n + 1 ≥ n n + 1
Equality condition: all n.
Use the substitution method to use conditions: 1 1 + x i = a i x i = 1 a i − 1 x i = a 1 + a 2 + ⋯ + a n + 1 a i − 1 x i = a 1 + a 2 + ⋯ + a i − 1 + a i + 1 + ⋯ + a n + 1 a i ≥ n a 1 a 2 ⋯ a i − 1 a i + 1 ⋯ a n + 1 n a i ( i = 1 , 2 , 3 , ⋯ , n + 1 ) \begin{gathered} \frac{1}{1+x_i}=a_i\\ x_i=\frac{1}{a_i}-1\\ x_i=\frac{a_1+a_2+\cdots+a_{n+1}}{a_i}-1\\ x_i=\frac{a_1+a_2+\cdots+a_{i-1}+a_{i+1}+\cdots+a_{n+1}}{a_i}\\ \ge\frac{n\sqrt[n]{a_1a_2\cdots a_{i-1}a_{i+1}\cdots a_{n+1}}}{a_i}(i=1,2,3,\cdots,n+1) \end{gathered} 1 + x i 1 = a i x i = a i 1 − 1 x i = a i a 1 + a 2 + ⋯ + a n + 1 − 1 x i = a i a 1 + a 2 + ⋯ + a i − 1 + a i + 1 + ⋯ + a n + 1 ≥ a i n n a 1 a 2 ⋯ a i − 1 a i + 1 ⋯ a n + 1 ( i = 1 , 2 , 3 , ⋯ , n + 1 )
The cumulative multiplication of all is proved