E:26460: Difference between revisions
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''' | '''26460 (Nicolae Mușuroia, Baia Mare)''' | ||
''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.''' | '''Soluție.''' | ||
Revision as of 12:42, 19 January 2025
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 . 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.}
