Question

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

$Dac\breve{a}\ a_1,\ a_2,\ \ldots\ ,\ a_n\in(0,+\infty),\ iar\ a_1\cdot a_2\cdot\ldots\cdot a_n=1\ s\breve{a}\ se\ arate\ c\breve{a}:$
$(a_{1} + \frac{a_{2} }{a_{3} } )^{2} + (a_{2} + \frac{a_{3} }{a_{4}} ) ^{2} + ... + (a_{n} + \frac{a_{1} }{a_{2} } )^{2} \geqslant 4n.$
Ion Tiotoi, profesor, Constanța

Answer 1

Aplicand inegalitatea Cauchy-Buniakovski obtinem:

$(\:a_{1}+\frac{a_{2}}{a_{3}})^2+(a_{2}+\frac{a_{3}}{a_{4}})^2+...+(a_{n}+\frac{a_{1}}{a_{2}})^2= \\ \\=\frac{1}{n} \cdot (1^2+1^2+...+1^2)[(a_{1}+\frac{a_{2}}{a_{3}})^2+...+(a_{0}+\frac{a_{1}}{a_{2}})^2] \geqslant \\ \\ \geqslant \frac{1}{n} \cdot [1 \cdot (a_{1}+\frac{a_{2}}{a_{3}})+1 \cdot (a_{2}+ \frac{a_{3}}{a_{4}})+...+1 \cdot (a_{n} + \frac{a_{1}}{a_{2}})]^2,~~~~~~~~(1). \\ \\ Dar~(a_{1}+\frac{a_{2}}{a_{3}})+...+(a_{n}+\frac{a_{1}}{a_{2}} \geqslant n ~\sqrt[n] {(a_{1}+\frac{a_{2}}{a_{3}}) \cdot (a_{2}+\frac{a_{3}}{a_{4}}) \cdot.... \cdot (a_{n} +\frac{a_{1}}{a_{2}})} \geqslant \\ \\ ~~~~~~\geqslant n \:\sqrt[n]{2 ~ \sqrt{a_{1} \cdot \frac{a_{2}}{a_{3}} \cdot... \cdot 2 ~ \sqrt{a_{n} \cdot \frac{a_{1}}{a_{2}}}}}= \\ \\ n \: \sqrt[n]{2^n \cdot \sqrt{a_{1}} \cdot a_{2} \cdot...\cdot a_{n} \cdot \frac{a_{1}}{a_{2}} \cdot \frac{a_{2}}{a_{3}} \cdot.... \cdot \frac{a_{n}} {a_{1}}}=n \: \sqrt[n] {2^n}=n \cdot 2,~~~~~(2),$
conform inegalitatii mediilor, aplicata de doua ori. Din (1) si (2) rezulta ca:
$(a_{1} + \frac{a_{2}}{a_{3}})^{2} + ... + (a _{n} + \frac{a_{1}}{a_{2} })^{2} \geqslant \frac{1}{n} \cdot (n \cdot 2) ^{2} = 4n ,~ceea \: ce \: trebuia \: demonstrat.$