Problema #12 – Inegalitatea Cauchy-Buniakovski-Schwarz

Să se demonstreze:

\frac{\sqrt[]{2}}{2}<\frac{\sqrt[]{1}+\sqrt[]{2}+...+\sqrt[]{2021}}{2021}\leq506

Demonstrăm întai a doua inegalitate:

\frac{\sqrt[]{1}+\sqrt[]{2}+…+\sqrt[]{2021}}{2021}\leq506

Din inegalitatea Cauchy-Buniakovski-Schwarz știm că:

(a_{1}\cdot b_{1}+a_{2}\cdot b_{2}+...+a_{n}\cdot b_{n})^2\leq\\\;\\\leq(a_{1}^2+a_{2}^2+...+a_{n}^2)\cdot\\\;\\\cdot(b_{1}^2+ b_{2}^2+...+ b_{n}^2)

Dacă bi este 1 pentru oricare i inegalitatea devine:

(a_{1}\cdot 1+a_{2}\cdot1+...+a_{n}\cdot 1)^2\leq\\\;\\\leq(a_{1}^2+a_{2}^2+...+a_{n}^2)\cdot\\\;\\\cdot\underbrace{(1^2+ 1^2+...+ 1^2)}_{n} \implies\\\;\\ \implies (a_{1}+a_{2}+...+a_{n})^2\leq\\\;\\\leq n \cdot(a_{1}^2+a_{2}^2+...+a_{n}^2)
Fie\;\; S=\sqrt[]{1}+\sqrt[]{2}+…+\sqrt[]{2021} \implies \\\;\\ \implies S^2 \leq 2021 \cdot((\sqrt[]{1})^2+(\sqrt[]{2})^2+...+(\sqrt[]{2021})^2) \implies \\\;\\ \implies S^2 \leq 2021 \cdot \underbrace{ (1+2+...+2021)}_{cf\;Gauss\;\frac{2021\cdot 2022}{2}} \implies \\\;\\\\ \implies S^2 \leq2021 \cdot \frac{2021\cdot 2022}{2}\implies \\\;\\\\ \implies S^2 \leq2021^2 \cdot1011 \implies \\\;\\
\left.
    \begin{array}{ll} 
\implies  S \leq2021 \cdot\sqrt[]{1011}\\\;\\\;\;\;\;\;\;\;\;\;\;\;\;\;\sqrt[]{1011}<506\end{array}
\right \}  
\implies\;\\\; \\ \implies \frac{S}{2021}<506 \implies \\\;\\\\ \implies\frac{\sqrt[]{1}+\sqrt[]{2}+…+\sqrt[]{2021} }{2021}<506

Demonstrăm prima inegalitate:

\frac{\sqrt[]{2}}{2}<\frac{\sqrt[]{1}+\sqrt[]{2}+...+\sqrt[]{2021}}{2021} \iff \\\;\\ \iff \frac{2021}{2}<\frac{\sqrt[]{1}+\sqrt[]{2}+...+\sqrt[]{2021}}{\sqrt[]{2}} \iff \\\;\\ \iff \frac{2021}{2}<\frac{\sqrt[]{2} \cdot(\sqrt[]{1}+\cdot\sqrt[]{2}+...+\cdot\sqrt[]{2021})}{2}  \iff \\\;\\ \iff 2021  < \sqrt[]{2} \cdot\underbrace{(\sqrt[]{1}+\cdot\sqrt[]{2}+...+\cdot\sqrt[]{2021})}_{>1\cdot2021} \iff \\\;\\\ \iff  2021<\sqrt[]{2}\cdot 2021 \;(A)\; q.e.d

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