Question

$Fie~a,b,c~apartin~[2,infinit).Demonstrati~ca: \\ \frac{ab+c}{c+1} + \frac{bc+a}{a+1} + \frac{ac+b}{b+1} \geq \frac{54}{a+b+c+3}$
Multumesc!

Answer 1

$\sum\frac{ab+c}{c+1}=\sum\frac{(\sqrt{ab+c})^2}{c+1}\ge\frac{(\sum \sqrt{ab+c})^2}{a+b+c+3}\ge\frac{(3\sqrt{6})^2}{a+b+c+3}=\frac{54}{a+b+c+3}.$

Answer 2

[tex]Aplicam\ inegalitatea\ C-B-S:\\ (\sum\frac{ab+c}{c+1})(a+b+c+3) \geq (\sqrt{ab+c}+\sqrt{ac+b}+\sqrt{bc+a})^2\\ \\ \sum \frac{ ab+c}{c+1}\geq \frac{(\sqrt{ab+c}+\sqrt{ac+b}+\sqrt{bc+a})^2}{a+b+c+3}\\ Dar\ stim\ ca: ab+c \geq6\Rightarrow\sqrt{ab+c}\geq \sqrt6(analog\ si\ pentru\\ \sqrt{ac+b}\ si \sqrt{bc+a})\\ Asadar:\\ \sum \frac{ab+c}{c+1}\geq \frac{(\sqrt6+\sqrt6+\sqrt6)^2}{a+b+c+3}\geq\frac{(3\sqrt6)^2}{a+b+c+3}\geq\frac{54}{a+b+c+3}\\ Q.E.D.[/tex]