One day I was trying to prove the Cauchy Schwarz inequality using the rearrangement inequality. However, at last I succeeded in proving it. Assume that a 1 , a 2 , … a n a_1,a_2,\dots a_n a 1 , a 2 , … a n and b 1 , b 2 , … b n b_1,b_2, \dots b_n b 1 , b 2 , … b n are sequences of real numbers. Here are my steps:
Step 1: Let A = ∑ i = 1 n a i 2 ≠ 0 , B = ∑ i = 1 n b i 2 ≠ 0 A=\displaystyle\sum_{i=1}^n a_i^2 \neq 0 \ , \ B=\displaystyle\sum_{i=1}^n b_i^2 \neq 0 A = i = 1 ∑ n a i 2 = 0 , B = i = 1 ∑ n b i 2 = 0
Step 2: 2 = 1 + 1 \ 2 = 1+1 2 = 1 + 1
Step 3: 2 = ∑ i = 1 n a i 2 A + ∑ i = 1 n b i 2 B \ 2=\dfrac{\displaystyle\sum_{i=1}^n a_i^2}{A} + \dfrac{\displaystyle\sum_{i=1}^n b_i^2}{B} 2 = A i = 1 ∑ n a i 2 + B i = 1 ∑ n b i 2
Step 4: 2 = a 1 2 A + a 2 2 A + … a n 2 A + b 1 2 B + b 2 2 B + … b n 2 B \ 2 = \dfrac{a_1^2}{A} +\dfrac{a_2^2}{A} + \dots \dfrac{a_n^2}{A} +\dfrac{b_1^2}{B} +\dfrac{b_2^2}{B} + \dots \dfrac{b_n^2}{B} 2 = A a 1 2 + A a 2 2 + … A a n 2 + B b 1 2 + B b 2 2 + … B b n 2
Step 5: So by rearrangement inequality, we have :
2 ≥ a 1 b 1 A B + a 2 b 2 A B + ⋯ + a n b n A B + b 1 a 1 A B + b 2 a 2 A B + ⋯ + b n a n A B 2 \geq \dfrac{a_1b_1}{\sqrt{AB}}+\dfrac{a_2b_2}{\sqrt{AB}} + \dots + \dfrac{a_nb_n}{\sqrt{AB}}+\dfrac{b_1a_1}{\sqrt{AB}}+\dfrac{b_2a_2}{\sqrt{AB}}+ \dots +\dfrac{b_na_n}{\sqrt{AB}} 2 ≥ A B a 1 b 1 + A B a 2 b 2 + ⋯ + A B a n b n + A B b 1 a 1 + A B b 2 a 2 + ⋯ + A B b n a n
Step 6: 2 A B ≥ 2 ( a 1 b 1 + a 2 b 2 + ⋯ + a n b n ) \ 2\sqrt{AB} \geq 2(a_1b_1+a_2b_2 + \dots + a_nb_n) 2 A B ≥ 2 ( a 1 b 1 + a 2 b 2 + ⋯ + a n b n )
Step 7: A B ≥ ( a 1 b 1 + a 2 b 2 + ⋯ + a n b n ) \ \sqrt{AB} \geq (a_1b_1+a_2b_2 + \dots + a_nb_n) A B ≥ ( a 1 b 1 + a 2 b 2 + ⋯ + a n b n )
Step 8: Squaring both sides, we have:
A B ≥ ( a 1 b 1 + a 2 b 2 … a n b n ) 2 AB \geq (a_1b_1+a_2b_2 \dots a_nb_n)^2 A B ≥ ( a 1 b 1 + a 2 b 2 … a n b n ) 2
Step 9: By replacing A,B we restore the form and we get the famous Cauchy Schwarz Inequality:
( ∑ i = 1 n a i 2 ) ( ∑ i = 1 n b i 2 ) ≥ ( ∑ i = 1 n a i b i ) 2 \left(\displaystyle\sum_{i=1}^n a_i^2\right)\left(\displaystyle\sum_{i=1}^n b_i^2\right) \geq \left(\displaystyle\sum_{i=1}^n a_ib_i \right)^2 ( i = 1 ∑ n a i 2 ) ( i = 1 ∑ n b i 2 ) ≥ ( i = 1 ∑ n a i b i ) 2
If you think I made a mistake in the proof, click the step number where I made a mistake. If you think that my proof is valid, click 0 0 0 .