E:26460

26460 (Nicolae Mușuroia, Baia Mare)

Să se arate că dacă 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} 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 ${\frac {1}{bc}}+{\frac {1}{ca}}+{\frac {1}{ab}}=1$. Avem

$$1+a^{2}=a^{2}\left(1+{\frac {1}{a^{2}}}\right)=a^{2}\left({\frac {1}{bc}}+{\frac {1}{ca}}+{\frac {1}{ab}}+{\frac {1}{a^{2}}}\right)=$$

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^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.}