Probleme de matematică Problema #12 – Inegalitatea Cauchy-Buniakovski-Schwarz ianuarie 20, 2021iunie 17, 2021 Anca Să se demonstreze: 22<1+2+...+20212021≤506\frac{\sqrt[]{2}}{2}<\frac{\sqrt[]{1}+\sqrt[]{2}+...+\sqrt[]{2021}}{2021}\leq50622<20211+2+...+2021≤506 Demonstrăm întai a doua inegalitate: 1+2+…+20212021≤506\frac{\sqrt[]{1}+\sqrt[]{2}+…+\sqrt[]{2021}}{2021}\leq50620211+2+…+2021≤506 Din inegalitatea Cauchy-Buniakovski-Schwarz știm că: (a1⋅b1+a2⋅b2+...+an⋅bn)2≤ ≤(a12+a22+...+an2)⋅ ⋅(b12+b22+...+bn2)(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)(a1⋅b1+a2⋅b2+...+an⋅bn)2≤≤(a12+a22+...+an2)⋅⋅(b12+b22+...+bn2) Dacă bi este 1 pentru oricare i inegalitatea devine: (a1⋅1+a2⋅1+...+an⋅1)2≤ ≤(a12+a22+...+an2)⋅ ⋅(12+12+...+12)⏟n ⟹ ⟹ (a1+a2+...+an)2≤ ≤n⋅(a12+a22+...+an2)(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)(a1⋅1+a2⋅1+...+an⋅1)2≤≤(a12+a22+...+an2)⋅⋅n(12+12+...+12)⟹⟹(a1+a2+...+an)2≤≤n⋅(a12+a22+...+an2) Fie S=1+2+…+2021 ⟹ ⟹ S2≤2021⋅((1)2+(2)2+...+(2021)2) ⟹ ⟹ S2≤2021⋅(1+2+...+2021)⏟cf Gauss 2021⋅20222 ⟹ ⟹ S2≤2021⋅2021⋅20222 ⟹ ⟹ S2≤20212⋅1011 ⟹ ⟹ S≤2021⋅1011 1011<506} ⟹ ⟹ S2021<506 ⟹ ⟹ 1+2+…+20212021<506Fie\;\; 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}<506FieS=1+2+…+2021⟹⟹S2≤2021⋅((1)2+(2)2+...+(2021)2)⟹⟹S2≤2021⋅cfGauss22021⋅2022(1+2+...+2021)⟹⟹S2≤2021⋅22021⋅2022⟹⟹S2≤20212⋅1011⟹⟹S≤2021⋅10111011<506⎭⎬⎫⟹⟹2021S<506⟹⟹20211+2+…+2021<506 Demonstrăm prima inegalitate: 22<1+2+...+20212021 ⟺ ⟺ 20212<1+2+...+20212 ⟺ ⟺ 20212<2⋅(1+⋅2+...+⋅2021)2 ⟺ ⟺ 2021<2⋅(1+⋅2+...+⋅2021)⏟>1⋅2021 ⟺ ⟺ 2021<2⋅2021 (A) q.e.d\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.d22<20211+2+...+2021⟺⟺22021<21+2+...+2021⟺⟺22021<22⋅(1+⋅2+...+⋅2021)⟺⟺2021<2⋅>1⋅2021(1+⋅2+...+⋅2021)⟺ ⟺2021<2⋅2021(A)q.e.d Pe aceeași temă Probleme de matematică Problemă rezolvată #42 – geometrie problemă cu romb aprilie 17, 2024aprilie 17, 2024 Anca Probleme de matematică Problemă rezolvată #41 – pregătire evaluare națională să se găsească numărul xyz martie 26, 2024martie 26, 2024 Anca
