E:26460: Difference between revisions

Latest revision as of 12:42, 19 January 2025

26460 (Nicolae Mușuroia)

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 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) = } 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.}

@@ Line 1: / Line 1: @@
-'''Problema 26460 (Nicolae Mușuroia, Baia Mare)'''
+'''26460 (Nicolae Mușuroia)'''
-''Să se arate că dacă a, b, c sunt numere reale strict pozitive cu <math>a + b + c = abc</math>, atunci <math>(1 + a^2)(1 + b^2)(1 + c^2) \geq 64</math>.''
+''Să se arate că dacă <math>a, b, c</math> sunt numere reale strict pozitive cu <math>a + b + c = abc</math>, atunci <math>(1 + a^2)(1 + b^2)(1 + c^2) \geq 64</math>.''
 '''Soluție.'''
-Relația <math>a + b + c = abc</math> se scrie <math>\frac{1}{bc} + \frac{1}{ca} + \frac{1}{ab} = 1</math>. Avem:
+Relația <math>a + b + c = abc</math> se scrie <math>\frac{1}{bc} + \frac{1}{ca} + \frac{1}{ab} = 1</math>. Avem<math display="block">
-<math>
 + 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) =
-</math>
+</math><math display="block">
-<math>
+= 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}}.
-= a^2 \left( \frac{1}{a} + \frac{1}{b} \right) \left( \frac{1}{b} + \frac{1}{c} \right) = \frac{(a + b)(a + c)}{\sqrt{bc}} \geq \frac{4a}{\sqrt{bc}}.
+</math>Deci <math display="block">(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.</math>
-</math>
-Deci <math>(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</math>.