Am I right or wrong?

Algebra Level 5

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 a1,a2,ana_1,a_2,\dots a_n and b1,b2,bnb_1,b_2, \dots b_n are sequences of real numbers. Here are my steps:

Step 1: Let A=i=1nai20 , B=i=1nbi20A=\displaystyle\sum_{i=1}^n a_i^2 \neq 0 \ , \ B=\displaystyle\sum_{i=1}^n b_i^2 \neq 0

Step 2:  2=1+1\ 2 = 1+1

Step 3:  2=i=1nai2A+i=1nbi2B\ 2=\dfrac{\displaystyle\sum_{i=1}^n a_i^2}{A} + \dfrac{\displaystyle\sum_{i=1}^n b_i^2}{B}

Step 4:  2=a12A+a22A+an2A+b12B+b22B+bn2B\ 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}

Step 5: So by rearrangement inequality, we have :

2a1b1AB+a2b2AB++anbnAB+b1a1AB+b2a2AB++bnanAB2 \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}}

Step 6:  2AB2(a1b1+a2b2++anbn)\ 2\sqrt{AB} \geq 2(a_1b_1+a_2b_2 + \dots + a_nb_n)

Step 7:  AB(a1b1+a2b2++anbn)\ \sqrt{AB} \geq (a_1b_1+a_2b_2 + \dots + a_nb_n)

Step 8: Squaring both sides, we have:

AB(a1b1+a2b2anbn)2AB \geq (a_1b_1+a_2b_2 \dots a_nb_n)^2

Step 9: By replacing A,B we restore the form and we get the famous Cauchy Schwarz Inequality:

(i=1nai2)(i=1nbi2)(i=1naibi)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

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 00.

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