mateforum.ro

a) Fie \( n\in\mathbb{N},n\geq2 \) si \( x_{1},y_{1},x_{2},y_{2},\ldots,x_{n},y_{n} \) numere reale strict pozitive. Sa se demonstreze ca daca \( x_{1}+x_{2}+\ldots+x_{n}\geq x_{1}y_{1}+x_{2}y_{2}+\ldots+x_{n}y_{n} \) atunci \( x_{1}+x_{2}+\ldots+x_{n}\leq\frac{x_{1}}{y_{1}}+\frac{x_{2}}{y_{2}}+\ldots+\frac{x_{n}}{y_{n}} \);
b) Folosind, eventual, punctul a), sa se demonstreze ca daca \( a,b,c \) sunt numere reale strict pozitive cu proprietatea ca \( ab+bc+ca \leq 3abc \), atunci \( a^3+b^3+c^3 \geq a+b+c \).

Romeo Ilie, O.N.M. 1999

a) \( x_1+x_2+\ldots +x_n\ \ge\ x_1y_1+x_2y_2+\ldots +x_ny_n\ (\ast) \)

Avem : \( \sum_{k=1}^n\frac{x_k}{y_k}=\sum_{k=1}^n \frac{x_k^2}{x_ky_k}\ \ge^{\small{CBS}}\ \frac{(x_1+x_2+\ldots +x_n)^2}{x_1y_1+x_2y_2+\ldots + x_ny_n}\ \ge^{(\ast)}\ \frac{(x_1+x_2+\ldots +x_n)^2}{x_1+x_2+\ldots +x_n}=x_1+x_2+\ldots +x_n \) .

b) Din \( ab+bc+ca\le 3abc\ \Longrightarrow\ \frac 1a+\frac 1b+\frac 1c\ \le\ 3\ (\ast\ast) \) .

Deoarece \( (a+b+c)\(\frac 1a+\frac 1b+\frac 1c\)\ \ge\ 9\ \Longrightarrow\ a+b+c\ \ge\ \frac{9}{\frac 1a+\frac 1b+\frac 1c}\ \ge^{(\ast\ast)}\ \frac 1a+\frac 1b+\frac 1c\ \ \ (\ast\ast\ast) \) .

Insa \( \(a\sqrt a\cdot \frac{1}{\sqrt a}+b\sqrt b\cdot\frac{1}{\sqrt b}+c\sqrt{c}\cdot\frac{1}{\sqrt c}\)^2\ \stackrel{\small{CBS}}{\le}\ (a^3+b^3+c^3)\(\frac 1a+\frac 1b+\frac 1c\)\ \le^{(\ast\ast\ast)}\ (a^3+b^3+c^3)(a+b+c) \)

Asadar, \( (a+b+c)^2\ \le\ (a^3+b^3+c^3)(a+b+c)\ \Longleftrightarrow\ a+b+c\ \le\ a^3+b^3+c^3 \) .

\( ab+bc+ca\le 3abc\ (1) \)
Din Inegalitatea lui Newton avem:
\( ab+bc+ca\ge sqrt{3abc(a+b+c)}\stackrel{(1)}\Longrightarrow 3abc\ge sqrt{3abc(a+b+c)}\Longrightarrow 3abc\ge a+b+c. \)
Ramane de demonstrat ca \( \sum a^3\ge 3abc \).
Dar \( \sum a^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\ge 0 \) q.e.d