Problema 21317*(G.M.1/1988)


Ion Tiotoi, profesor, Constanța
GreenEyes71:
Îți ofer o indicație completă de rezolvare. Vei avea de aplicat de 3 ori inegalitatea mediilor (media aritmetică este mai mare sau egală decât cea geometrică), pentru 3 sume, fiecare sumă are n termeni.
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Aplicand inegalitatea Cauchy-Buniakovski obtinem:
![(\:a_{1}+\frac{a_{2}}{a_{3}})^2+(a_{2}+\frac{a_{3}}{a_{4}})^2+...+(a_{n}+\frac{a_{1}}{a_{2}})^2= \\ \\=\frac{1}{n} \cdot (1^2+1^2+...+1^2)[(a_{1}+\frac{a_{2}}{a_{3}})^2+...+(a_{0}+\frac{a_{1}}{a_{2}})^2] \geqslant \\ \\ \geqslant \frac{1}{n} \cdot [1 \cdot (a_{1}+\frac{a_{2}}{a_{3}})+1 \cdot (a_{2}+ \frac{a_{3}}{a_{4}})+...+1 \cdot (a_{n} + \frac{a_{1}}{a_{2}})]^2,~~~~~~~~(1). \\ \\ Dar~(a_{1}+\frac{a_{2}}{a_{3}})+...+(a_{n}+\frac{a_{1}}{a_{2}} \geqslant n ~\sqrt[n] {(a_{1}+\frac{a_{2}}{a_{3}}) \cdot (a_{2}+\frac{a_{3}}{a_{4}}) \cdot.... \cdot (a_{n} +\frac{a_{1}}{a_{2}})} \geqslant \\ \\ ~~~~~~\geqslant n \:\sqrt[n]{2 ~ \sqrt{a_{1} \cdot \frac{a_{2}}{a_{3}} \cdot... \cdot 2 ~ \sqrt{a_{n} \cdot \frac{a_{1}}{a_{2}}}}}= \\ \\ n \: \sqrt[n]{2^n \cdot \sqrt{a_{1}} \cdot a_{2} \cdot...\cdot a_{n} \cdot \frac{a_{1}}{a_{2}} \cdot \frac{a_{2}}{a_{3}} \cdot.... \cdot \frac{a_{n}} {a_{1}}}=n \: \sqrt[n] {2^n}=n \cdot 2,~~~~~(2), (\:a_{1}+\frac{a_{2}}{a_{3}})^2+(a_{2}+\frac{a_{3}}{a_{4}})^2+...+(a_{n}+\frac{a_{1}}{a_{2}})^2= \\ \\=\frac{1}{n} \cdot (1^2+1^2+...+1^2)[(a_{1}+\frac{a_{2}}{a_{3}})^2+...+(a_{0}+\frac{a_{1}}{a_{2}})^2] \geqslant \\ \\ \geqslant \frac{1}{n} \cdot [1 \cdot (a_{1}+\frac{a_{2}}{a_{3}})+1 \cdot (a_{2}+ \frac{a_{3}}{a_{4}})+...+1 \cdot (a_{n} + \frac{a_{1}}{a_{2}})]^2,~~~~~~~~(1). \\ \\ Dar~(a_{1}+\frac{a_{2}}{a_{3}})+...+(a_{n}+\frac{a_{1}}{a_{2}} \geqslant n ~\sqrt[n] {(a_{1}+\frac{a_{2}}{a_{3}}) \cdot (a_{2}+\frac{a_{3}}{a_{4}}) \cdot.... \cdot (a_{n} +\frac{a_{1}}{a_{2}})} \geqslant \\ \\ ~~~~~~\geqslant n \:\sqrt[n]{2 ~ \sqrt{a_{1} \cdot \frac{a_{2}}{a_{3}} \cdot... \cdot 2 ~ \sqrt{a_{n} \cdot \frac{a_{1}}{a_{2}}}}}= \\ \\ n \: \sqrt[n]{2^n \cdot \sqrt{a_{1}} \cdot a_{2} \cdot...\cdot a_{n} \cdot \frac{a_{1}}{a_{2}} \cdot \frac{a_{2}}{a_{3}} \cdot.... \cdot \frac{a_{n}} {a_{1}}}=n \: \sqrt[n] {2^n}=n \cdot 2,~~~~~(2),](https://tex.z-dn.net/?f=%28%5C%3Aa_%7B1%7D%2B%5Cfrac%7Ba_%7B2%7D%7D%7Ba_%7B3%7D%7D%29%5E2%2B%28a_%7B2%7D%2B%5Cfrac%7Ba_%7B3%7D%7D%7Ba_%7B4%7D%7D%29%5E2%2B...%2B%28a_%7Bn%7D%2B%5Cfrac%7Ba_%7B1%7D%7D%7Ba_%7B2%7D%7D%29%5E2%3D+%5C%5C+%5C%5C%3D%5Cfrac%7B1%7D%7Bn%7D+%5Ccdot+%281%5E2%2B1%5E2%2B...%2B1%5E2%29%5B%28a_%7B1%7D%2B%5Cfrac%7Ba_%7B2%7D%7D%7Ba_%7B3%7D%7D%29%5E2%2B...%2B%28a_%7B0%7D%2B%5Cfrac%7Ba_%7B1%7D%7D%7Ba_%7B2%7D%7D%29%5E2%5D+%5Cgeqslant+%5C%5C+%5C%5C+%5Cgeqslant+%5Cfrac%7B1%7D%7Bn%7D+%5Ccdot+%5B1+%5Ccdot+%28a_%7B1%7D%2B%5Cfrac%7Ba_%7B2%7D%7D%7Ba_%7B3%7D%7D%29%2B1+%5Ccdot++%28a_%7B2%7D%2B+%5Cfrac%7Ba_%7B3%7D%7D%7Ba_%7B4%7D%7D%29%2B...%2B1+%5Ccdot+%28a_%7Bn%7D+%2B+%5Cfrac%7Ba_%7B1%7D%7D%7Ba_%7B2%7D%7D%29%5D%5E2%2C%7E%7E%7E%7E%7E%7E%7E%7E%281%29.+%5C%5C+%5C%5C+Dar%7E%28a_%7B1%7D%2B%5Cfrac%7Ba_%7B2%7D%7D%7Ba_%7B3%7D%7D%29%2B...%2B%28a_%7Bn%7D%2B%5Cfrac%7Ba_%7B1%7D%7D%7Ba_%7B2%7D%7D+%5Cgeqslant+n+%7E%5Csqrt%5Bn%5D+%7B%28a_%7B1%7D%2B%5Cfrac%7Ba_%7B2%7D%7D%7Ba_%7B3%7D%7D%29+%5Ccdot+%28a_%7B2%7D%2B%5Cfrac%7Ba_%7B3%7D%7D%7Ba_%7B4%7D%7D%29+%5Ccdot....+%5Ccdot+%28a_%7Bn%7D+%2B%5Cfrac%7Ba_%7B1%7D%7D%7Ba_%7B2%7D%7D%29%7D+%5Cgeqslant+%5C%5C+%5C%5C+%7E%7E%7E%7E%7E%7E%5Cgeqslant+n+%5C%3A%5Csqrt%5Bn%5D%7B2+%7E+%5Csqrt%7Ba_%7B1%7D+%5Ccdot+%5Cfrac%7Ba_%7B2%7D%7D%7Ba_%7B3%7D%7D+%5Ccdot...+%5Ccdot+2+%7E+%5Csqrt%7Ba_%7Bn%7D+%5Ccdot+%5Cfrac%7Ba_%7B1%7D%7D%7Ba_%7B2%7D%7D%7D%7D%7D%3D+%5C%5C+%5C%5C+n+%5C%3A+%5Csqrt%5Bn%5D%7B2%5En+%5Ccdot+%5Csqrt%7Ba_%7B1%7D%7D++%5Ccdot+a_%7B2%7D+%5Ccdot...%5Ccdot+a_%7Bn%7D+%5Ccdot+%5Cfrac%7Ba_%7B1%7D%7D%7Ba_%7B2%7D%7D+%5Ccdot+%5Cfrac%7Ba_%7B2%7D%7D%7Ba_%7B3%7D%7D+%5Ccdot....+%5Ccdot+%5Cfrac%7Ba_%7Bn%7D%7D+%7Ba_%7B1%7D%7D%7D%3Dn+%5C%3A+%5Csqrt%5Bn%5D+%7B2%5En%7D%3Dn+%5Ccdot+2%2C%7E%7E%7E%7E%7E%282%29%2C)
conform inegalitatii mediilor, aplicata de doua ori. Din (1) si (2) rezulta ca:

conform inegalitatii mediilor, aplicata de doua ori. Din (1) si (2) rezulta ca:
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