E:26460

Problema 26460 (Nicolae Mușuroia, Baia Mare)

Să se arate că dacă a, b, c sunt numere reale strict pozitive cu Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a + b + c = abc} , atunci Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (1 + a^2)(1 + b^2)(1 + c^2) \geq 64} .

Soluție.

Relația Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a + b + c = abc} se scrie Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{bc} + \frac{1}{ca} + \frac{1}{ab} = 1} . AvemFailed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1 + a^2 = a^2 \left( 1 + \frac{1}{a^2} \right) = a^2 \left( \frac{1}{bc} + \frac{1}{ca} + \frac{1}{ab} + \frac{1}{a^2} \right) = }

$$=a^{2}\left({\frac {1}{a}}+{\frac {1}{b}}\right)\left({\frac {1}{a}}+{\frac {1}{c}}\right)={\frac {(a+b)(a+c)}{bc}}\geq {\frac {4a}{\sqrt {bc}}}.$$

Deci Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (1 + a^2)(1 + b^2)(1 + c^2) \geq \frac{4a}{\sqrt{bc}} \cdot \frac{4b}{\sqrt{ac}} \cdot \frac{4c}{\sqrt{ab}} = 64.}