Question

Sa se demonstreze ca x=$\frac{a+b}{c}$+c($\frac{1}{a} + \frac{1}{b}$)$\geq 2$, oricare ar fi a,b,c≥0

Answer 1

$\frac{a+b}{c}+c( \frac{1}{a} + \frac{1}{b}) \geq 2 \\ \\ Inmultim~cu~abc~ambii~membri,~si~obtinem: \\ \\ ab(a+b)+abc^2( \frac{1}{a}+ \frac{1}{b}) \geq 2abc \Leftrightarrow \\ \\ \Leftrightarrow ab(a+b) +ab c^2 \cdot \frac{a+b}{ab} \geq 2abc \Leftrightarrow \\ \\ \Leftrightarrow ab(a+b)+c^2(a+b) \geq 2abc \Leftrightarrow \\ \\ \Leftrightarrow(a+b)(ab+c^2) \geq 2abc. \\ \\ Utilizand~inegalitatile~dintre~media~aritmetica~si~geometrica,~ \\ \\ obtinem:$

$\frac{a+c}{2} \geq \sqrt{ac} \Rightarrow a+c \geq 2 \sqrt{ac}. \\ \\ \frac{ab+c^2}{2} \geq \sqrt{abc^2}=c \sqrt{ab} \Rightarrow ab+c^2 \geq 2c \sqrt{ab}. \\ \\ Din~a+c \geq 2 \sqrt{ac}~si~ab+c^2 \geq 2c \sqrt{ab}~rezulta~(a+c)(ab+c^2) \geq 2abc, \\ \\ adica~ceea~ce~trebuia~demonstrat~(Q.E.D.)$