Cauchy Schwarz is too useful...

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

Simplest form of it is

ac+bda2+b2c2+d2\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 {a1,a2,a3,...,ana_1,a_2,a_3,...,a_n } and {b1,b2,b3,...,bnb_1,b_2,b_3,...,b_n} , the following inequality holds :-

j=1najbj(j=1naj2)12(j=1nbj2)12\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}}


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

in solving problems, you have to chose your sets {aia_i} and {bib_i} SMARTLY\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,dR+a,b,c,d \in \mathbb{R} ^+ , prove the following -

1a+9b+25c+49d256a+b+c+d\displaystyle \dfrac{1}{a} +\dfrac{9}{b}+\dfrac{25}{c}+\dfrac{49}{d} \geq \dfrac{256}{a+b+c+d}

#Algebra #Inequalities #CauchySchwarzInequality #Learning

Note by Aditya Raut
6 years, 11 months ago

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Comments

(a2+b2+c2+d2)((1a)2+(3b)2+(5c)2+(7d)2)(a1a+b3b+c5c+d7d)2\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}

Samuraiwarm Tsunayoshi - 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 ^__^

Samuraiwarm Tsunayoshi - 6 years, 11 months ago

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

Imgur Imgur

Aditya Raut - 6 years, 11 months ago

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@Aditya Raut Thank you meow!!! ^__^

Samuraiwarm Tsunayoshi - 6 years, 10 months ago

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

hemang sarkar - 6 years, 11 months ago
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