home / maths / techniques / algebraic_techniques_04_lagrange_identity_and_cauchy_schwarz_inequality Algebraic Techniques 04 - Lagrange's Identity and Cauchy-Schwarz Inequality
✔ Lagrange’s Identity ✔ Cauchy-Schwarz Inequality ✔ Corollary of Cauchy-Schwarz Inequality ✔ Minkowski Inequality For all real numbers a \, a a , b b b , and c c c , prove the following equation. For real numbers a \, a a and b b b , prove the following inequality. For positive real numbers a \, a a , b b b , and c \, c c such that a 2 + b 2 + c 2 = 3 \, a^2 + b^2 + c^2 = 3 a 2 + b 2 + c 2 = 3 , prove the following inequality. Reference ✔ Lagrange’s Identity For all real numbers x 1 , ⋯ , x n , and y 1 , ⋯ , y n , ( x 1 2 + ⋯ + x n 2 ) ( y 1 2 + ⋯ + y n 2 ) = ( x 1 y 1 + ⋯ + x n y n ) 2 + ∑ 1 ≤ i < j ≤ n ( x i y j − x j y i ) 2 Especially, for real numbers a , b , c , d , ( a 2 + b 2 ) ( c 2 + d 2 ) = ( a c + b d ) 2 + ( a d − b c ) 2 = ( a c − b d ) 2 + ( a d + b c ) 2 \begin{aligned} &\text{For all real numbers} \; x_1, \cdots, x_n, \text{and} \; y_1, \cdots, y_n, \\\\ &\qquad (x_1^2 + \cdots + x_n^2)(y_1^2 + \cdots + y_n^2) = (x_1y_1 + \cdots + x_ny_n)^2 + \sum_{1 \leq i < j \leq n} (x_i y_j - x_j y_i)^2 \\\\\\ &\text{Especially, for real numbers} \; a, b, c, d, \\\\ &\qquad (a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2 = (ac - bd)^2 + (ad + bc)^2 \end{aligned} For all real numbers x 1 , ⋯ , x n , and y 1 , ⋯ , y n , ( x 1 2 + ⋯ + x n 2 ) ( y 1 2 + ⋯ + y n 2 ) = ( x 1 y 1 + ⋯ + x n y n ) 2 + 1 ≤ i < j ≤ n ∑ ( x i y j − x j y i ) 2 Especially, for real numbers a , b , c , d , ( a 2 + b 2 ) ( c 2 + d 2 ) = ( a c + b d ) 2 + ( a d − b c ) 2 = ( a c − b d ) 2 + ( a d + b c ) 2 ✔ Cauchy-Schwarz Inequality For all real numbers x 1 , ⋯ , x n , and y 1 , ⋯ , y n , ( x 1 2 + ⋯ + x n 2 ) ( y 1 2 + ⋯ + y n 2 ) ≥ ( x 1 y 1 + ⋯ + x n y n ) 2 The equal sign is valid when (1) x 1 = ⋯ = x n = 0 , or (2) there exists a real number t such that y i = t x i ( i = 1 , ⋯ , n ) \begin{aligned} &\text{For all real numbers} \; x_1, \cdots, x_n, \text{and} \; y_1, \cdots, y_n, \\\\ &\qquad\qquad (x_1^2 + \cdots + x_n^2)(y_1^2 + \cdots + y_n^2) \geq (x_1y_1 + \cdots + x_ny_n)^2 \\\\ &\text{The equal sign is valid when} \\ &\qquad\qquad \text{(1)} \; x_1 = \cdots = x_n = 0, \text{or}\\ &\qquad\qquad \text{(2) there exists a real number} \; t \; \text{such that} \; y_i = t x_i \; (i = 1, \cdots, n) \end{aligned} For all real numbers x 1 , ⋯ , x n , and y 1 , ⋯ , y n , ( x 1 2 + ⋯ + x n 2 ) ( y 1 2 + ⋯ + y n 2 ) ≥ ( x 1 y 1 + ⋯ + x n y n ) 2 The equal sign is valid when (1) x 1 = ⋯ = x n = 0 , or (2) there exists a real number t such that y i = t x i ( i = 1 , ⋯ , n ) ✔ Corollary of Cauchy-Schwarz Inequality For all real numbers a 1 , ⋯ , a n , and all positive real numbers b 1 , ⋯ , b n , a 1 2 b 1 + ⋯ + a n 2 b n ≥ ( a 1 + ⋯ + a n ) 2 b 1 + ⋯ + b n \begin{aligned} &\text{For all real numbers} \; a_1, \cdots, a_n, \text{and all positive real numbers} \; b_1, \cdots, b_n, \\\\ &\qquad\qquad\qquad\qquad \frac{a_1^2}{b_1} + \cdots + \frac{a_n^2}{b_n} \geq \frac{(a_1 + \cdots + a_n)^2}{b_1 + \cdots + b_n} \end{aligned} For all real numbers a 1 , ⋯ , a n , and all positive real numbers b 1 , ⋯ , b n , b 1 a 1 2 + ⋯ + b n a n 2 ≥ b 1 + ⋯ + b n ( a 1 + ⋯ + a n ) 2 ✔ Minkowski Inequality For all real numbers x 1 , ⋯ , x n , and y 1 , ⋯ , y n , x 1 2 + y 1 2 + ⋯ + x n 2 + y n 2 ≥ ( x 1 + ⋯ + x n ) 2 + ( y 1 + ⋯ + y n ) 2 \begin{aligned} &\text{For all real numbers} \; x_1, \cdots, x_n, \text{and} \; y_1, \cdots, y_n, \\\\ &\qquad\qquad \sqrt{x_1^2 + y_1^2} + \cdots + \sqrt{x_n^2 + y_n^2} \geq \sqrt{(x_1 + \cdots + x_n)^2 + (y_1 + \cdots + y_n)^2} \end{aligned} For all real numbers x 1 , ⋯ , x n , and y 1 , ⋯ , y n , x 1 2 + y 1 2 + ⋯ + x n 2 + y n 2 ≥ ( x 1 + ⋯ + x n ) 2 + ( y 1 + ⋯ + y n ) 2 For all real numbers a \, a a , b b b , and c c c , prove the following equation. ( 1 + a 2 ) ( 1 + b 2 ) ( 1 + c 2 ) = ( a + b + c − a b c ) 2 + ( a b + b c + c a − 1 ) 2 \begin{aligned} (1 + a^2)(1 + b^2)(1 + c^2) = (a + b + c - abc)^2 + (ab + bc + ca - 1)^2 \end{aligned} ( 1 + a 2 ) ( 1 + b 2 ) ( 1 + c 2 ) = ( a + b + c − ab c ) 2 + ( ab + b c + c a − 1 ) 2 [Solution] Applying Lagrange’s identity twice, ( 1 + a 2 ) ( 1 + b 2 ) = ( a + b ) 2 + ( 1 − a b ) 2 ⟹ ( 1 + a 2 ) ( 1 + b 2 ) ( 1 + c 2 ) = ( ( a + b ) 2 + ( 1 − a b ) 2 ) ( 1 + c 2 ) = ( a + b + c ( 1 − a b ) ) 2 + ( ( a + b ) c − ( 1 − a b ) ) 2 = ( a + b + c − a b c ) 2 + ( a b + b c + c a − 1 ) 2 \begin{aligned} (1 + a^2)(1 + b^2) &= (a + b)^2 + (1 - ab)^2 \\\\ \Longrightarrow (1 + a^2)(1 + b^2)(1 + c^2) &= ((a + b)^2 + (1 - ab)^2)(1 + c^2) \\ &= (a + b + c(1 - ab))^2 + ((a + b)c - (1 - ab))^2 \\ &= (a + b + c - abc)^2 + (ab + bc + ca - 1)^2 \end{aligned} ( 1 + a 2 ) ( 1 + b 2 ) ⟹ ( 1 + a 2 ) ( 1 + b 2 ) ( 1 + c 2 ) = ( a + b ) 2 + ( 1 − ab ) 2 = (( a + b ) 2 + ( 1 − ab ) 2 ) ( 1 + c 2 ) = ( a + b + c ( 1 − ab ) ) 2 + (( a + b ) c − ( 1 − ab ) ) 2 = ( a + b + c − ab c ) 2 + ( ab + b c + c a − 1 ) 2
For real numbers a \, a a and b b b , prove the following inequality. a 4 + b 4 ≥ a b ( a 2 + b 2 ) \begin{aligned} a^4 + b^4 \geq ab (a^2 + b^2) \end{aligned} a 4 + b 4 ≥ ab ( a 2 + b 2 ) [Solution] By Cauchy-Schwarz Inequality and AM-GM , ( 1 2 + 1 2 ) ( x 2 + y 2 ) ≥ ( x + y ) 2 , x 2 + y 2 2 ≥ x y \begin{aligned} (1^2 + 1^2)(x^2 + y^2) \geq (x + y)^2, \quad \frac{x^2 + y^2}{2} \geq xy \end{aligned} ( 1 2 + 1 2 ) ( x 2 + y 2 ) ≥ ( x + y ) 2 , 2 x 2 + y 2 ≥ x y
Therefore, a 4 + b 4 ≥ ( a 2 + b 2 ) 2 2 = a 2 + b 2 2 ( a 2 + b 2 ) ≥ a b ( a 2 + b 2 ) \begin{aligned} a^4 + b^4 \geq \frac{(a^2 + b^2)^2}{2} = \frac{a^2 + b^2}{2} (a^2 + b^2) \geq ab (a^2 + b^2) \end{aligned} a 4 + b 4 ≥ 2 ( a 2 + b 2 ) 2 = 2 a 2 + b 2 ( a 2 + b 2 ) ≥ ab ( a 2 + b 2 )
For positive real numbers a \, a a , b b b , and c \, c c such that a 2 + b 2 + c 2 = 3 \, a^2 + b^2 + c^2 = 3 a 2 + b 2 + c 2 = 3 , prove the following inequality. a 3 b 2 + b 3 c 2 + c 3 a 2 ≥ a 2 b + b 2 c + c 2 a \begin{aligned} \frac{a^3}{b^2} + \frac{b^3}{c^2} + \frac{c^3}{a^2} \geq \frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \end{aligned} b 2 a 3 + c 2 b 3 + a 2 c 3 ≥ b a 2 + c b 2 + a c 2 [Solution] By Cauchy-Schwarz Inequality, ( a 2 b + b 2 c + c 2 a ) 2 = ( a 3 b 2 a + b 3 c 2 b + c 3 a 2 c ) 2 ≤ ( a 3 b 2 + b 3 c 2 + c 3 a 2 ) ( a + b + c ) \begin{aligned} \left(\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a}\right)^2 = \left(\sqrt{\frac{a^3}{b^2}}\sqrt{a} + \sqrt{\frac{b^3}{c^2}}\sqrt{b} + \sqrt{\frac{c^3}{a^2}}\sqrt{c}\right)^2 \leq \left(\frac{a^3}{b^2} + \frac{b^3}{c^2} + \frac{c^3}{a^2}\right) (a + b + c) \end{aligned} ( b a 2 + c b 2 + a c 2 ) 2 = ( b 2 a 3 a + c 2 b 3 b + a 2 c 3 c ) 2 ≤ ( b 2 a 3 + c 2 b 3 + a 2 c 3 ) ( a + b + c )
It is sufficient to prove that a + b + c ≤ a 2 b + b 2 c + c 2 a \begin{aligned} a + b + c \leq \frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \end{aligned} a + b + c ≤ b a 2 + c b 2 + a c 2
By corollary of Cauchy-Schwarz Inequality, a 2 b + b 2 c + c 2 a ≥ ( a + b + c ) 2 a + b + c = a + b + c \begin{aligned} \frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \geq \frac{(a + b + c)^2}{a + b + c} = a + b + c \end{aligned} b a 2 + c b 2 + a c 2 ≥ a + b + c ( a + b + c ) 2 = a + b + c
Reference [1] Titu Andreescu, 105 Algebra Problems.
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