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 \(a_1,a_2,\dots a_n\) and \(b_1,b_2, \dots b_n\) are sequences of real numbers. Here are my steps:

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

**Step 2:** \(\ 2 = 1+1\)

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

**Step 4:** \(\ 2 = \dfrac{a_1^2}{A} +\dfrac{a_2^2}{A} + \dots \dfrac{a_n^2}{A} +\dfrac{b_1^2}{B} +\dfrac{b_2^2}{B} + \dots \dfrac{b_n^2}{B} \)

**Step 5:** So by rearrangement inequality, we have :

\(2 \geq \dfrac{a_1b_1}{\sqrt{AB}}+\dfrac{a_2b_2}{\sqrt{AB}} + \dots + \dfrac{a_nb_n}{\sqrt{AB}}+\dfrac{b_1a_1}{\sqrt{AB}}+\dfrac{b_2a_2}{\sqrt{AB}}+ \dots +\dfrac{b_na_n}{\sqrt{AB}}\)

**Step 6:** \(\ 2\sqrt{AB} \geq 2(a_1b_1+a_2b_2 + \dots + a_nb_n)\)

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

**Step 8:** Squaring both sides, we have:

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

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

\(\left(\displaystyle\sum_{i=1}^n a_i^2\right)\left(\displaystyle\sum_{i=1}^n b_i^2\right) \geq \left(\displaystyle\sum_{i=1}^n a_ib_i \right)^2\)

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 \(0\).

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