E:16902

E:16902 (Melania-Iulia Dobrican)

Fie numerele reale pozitive 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 x} , 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 y} , 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 xy=4} . Arătaţi că 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}{x+4} + \frac{1}{y+4} \le \frac{1}{3}.}

Soluție

Avem echivalenţele

$${\frac {1}{x+4}}+{\frac {1}{y+4}}\leq {\frac {1}{3}}\Leftrightarrow 3y+12+3x+12\leq xy+4x+4y+16\Leftrightarrow x+y\geq 8-xy.$$

Cum $xy=4$, rezultă $x+y\geq 4$. Au loc echivalenţele

$$x+y\geq 4\Leftrightarrow {\frac {x+y}{2}}\geq 2={\sqrt {4}}\Leftrightarrow {\frac {x+y}{2}}\geq {\sqrt {xy}},$$

inegalitate care are loc pentru orice numere reale pozitive $x$, $y$.

Cazul de egalitate are loc pentru $x=y=2$.