# Cauchy Schwarz is too useful...

The famous Cauchy Schwarz Inequality is a very very useful result in $\color{#D61F06}{\textbf{inequalities}}$

Simplest form of it is

$\displaystyle \left| ac+bd \right| \leq \sqrt{a^2+b^2} \sqrt{c^2+d^2}$,

and it's generalized form is

For any two sets of real numbers {$a_1,a_2,a_3,...,a_n$} and {$b_1,b_2,b_3,...,b_n$} , the following inequality holds :-

$\displaystyle \left| \sum_{j=1}^n a_jb_j \right| \leq \biggl( \sum_{j=1}^n a_j^2 \biggr)^{\frac{1}{2}} \biggl( \sum_{j=1}^n b_j^2 \biggr) ^{\frac{1}{2}}$

$\color{#3D99F6}{\textbf{What's important is that}}$

in solving problems, you have to chose your sets {$a_i$} and {$b_i$} $\color{#D61F06}{\textbf{S}} \color{#20A900}{\textbf{M}} \color{#3D99F6}{\textbf{A}}\color{#EC7300}{\textbf{R}}\color{#69047E}{\textbf{T}} \color{limegreen}{\textbf{L}}\color{#95D3FE}{\textbf{Y}}$ to reach your required answer...

For practice, here is one problem (Not made by me, but i changed the numericals than what I had seen). Try this, and it uses the result told above....

For $a,b,c,d \in \mathbb{R} ^+$, prove the following -

$\displaystyle \dfrac{1}{a} +\dfrac{9}{b}+\dfrac{25}{c}+\dfrac{49}{d} \geq \dfrac{256}{a+b+c+d}$ Note by Aditya Raut
6 years, 11 months ago

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## Comments

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$\displaystyle (\sqrt{a}^{2}+\sqrt{b}^{2}+\sqrt{c}^{2}+\sqrt{d}^{2})((\frac{1}{\sqrt{a}})^{2}+(\frac{3}{\sqrt{b}})^{2}+(\frac{5}{\sqrt{c}})^{2}+(\frac{7}{\sqrt{d}})^{2}) \geq (\sqrt{a}\frac{1}{\sqrt{a}}+\sqrt{b}\frac{3}{\sqrt{b}}+\sqrt{c}\frac{5}{\sqrt{c}}+\sqrt{d}\frac{7}{\sqrt{d}})^{2}$

- 6 years, 11 months ago

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lol it went through the space

Do you have any more problems for us to practice Cauchy Schwarz's inequality? Thank you ^__^

- 6 years, 11 months ago

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Yes my friend, I will post some problems rather than notes now :D Imgur

- 6 years, 11 months ago

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Thank you meow!!! ^__^

- 6 years, 11 months ago

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in general, we can just try to prove the Engel form of CS inequality. this will be proved automatically then.

- 6 years, 11 months ago

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