E:26460

Problema 26460 (Nicolae Mușuroia, Baia Mare)

Să se arate că dacă a, b, c sunt numere reale strict pozitive cu $a+b+c=abc$, atunci $(1+a^{2})(1+b^{2})(1+c^{2})\geq 64$.

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} . Avem

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

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

Deci

$$(1+a^{2})(1+b^{2})(1+c^{2})\geq {\frac {4a}{\sqrt {bc}}}\cdot {\frac {4b}{\sqrt {ac}}}\cdot {\frac {4c}{\sqrt {ab}}}=64.$$