The famous Cauchy Schwarz Inequality is a very very useful result in \(\color{Red}{\textbf{inequalities}}\)

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 :-

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

\(\color{Blue}{\textbf{What's important is that}}\)

in solving problems, you have to chose your sets {\(a_i\)} and {\(b_i\)} \(\color{Red}{\textbf{S}} \color{Green}{\textbf{M}} \color{Blue}{\textbf{A}}\color{Orange}{\textbf{R}}\color{Purple}{\textbf{T}} \color{LimeGreen}{\textbf{L}}\color{LightBlue}{\textbf{Y}}\) 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 -

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

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TopNewest\(\displaystyle (\sqrt{a}^{2}+\sqrt{b}^{2}+\sqrt{c}^{2}+\sqrt{d}^{2})((\frac{1}{\sqrt{a}})^{2}+(\frac{3}{\sqrt{b}})^{2}+(\frac{5}{\sqrt{c}})^{2}+(\frac{7}{\sqrt{d}})^{2}) \geq (\sqrt{a}\frac{1}{\sqrt{a}}+\sqrt{b}\frac{3}{\sqrt{b}}+\sqrt{c}\frac{5}{\sqrt{c}}+\sqrt{d}\frac{7}{\sqrt{d}})^{2}\) – Samuraiwarm Tsunayoshi · 2 years, 3 months ago

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Do you have any more problems for us to practice Cauchy Schwarz's inequality? Thank you ^__^ – Samuraiwarm Tsunayoshi · 2 years, 3 months ago

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– Samuraiwarm Tsunayoshi · 2 years, 3 months ago

Thank you meow!!! ^__^Log in to reply

in general, we can just try to prove the Engel form of CS inequality. this will be proved automatically then. – Hemang Sarkar · 2 years, 3 months ago

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