Question

AJUTOR !!!! MULTUMESC ANTICIPAT !!!

Answer 1

[tex]\displaystyle \\ \texttt{Demonstrati ca:} \\ \\\frac{2}{3}<(\sqrt{6})^{-1} +(\sqrt{15})^{-1} +(\sqrt{35})^{-1} <\frac{16}{15} \\ \\ \texttt{Rezolvare: } \\ \\\frac{2}{3} \ \textless \ \frac{1}{\sqrt{6}} + \frac{1}{\sqrt{15}} + \frac{1}{\sqrt{35}} \ \textless \ \frac{16}{15}[/tex]


[tex]\displaystyle \\ \frac{2}{3} \textless \frac{1}{\sqrt{6}} + \frac{1}{\sqrt{15}} + \frac{1}{\sqrt{35}} \textless \frac{16}{15} \\ \\ \texttt{Demonstram prima inegalitate: } \\ \\ \frac{1}{\sqrt{6}} \ \textgreater \ \frac{1}{\sqrt{9}}~~si ~~\frac{1}{\sqrt{15}} \ \textgreater \ \frac{1}{\sqrt{16}} ~~si ~~\frac{1}{\sqrt{35}} \ \textgreater \ \frac{1}{\sqrt{36}} \\ \\ \Longrightarrow ~~\frac{1}{\sqrt{6}} + \frac{1}{\sqrt{15}} + \frac{1}{\sqrt{35}} \ \textgreater \ \frac{1}{\sqrt{9}} + \frac{1}{\sqrt{16}} + \frac{1}{\sqrt{36}} =[/tex]


$\displaystyle \\ =\frac{1}{3} + \frac{1}{4} + \frac{1}{6} = \frac{4+3+2}{12}= \frac{9}{12}\ \textgreater \ \frac{8}{12}=\frac{2}{3} \\ \\ \Longrightarrow ~~\boxed{\frac{2}{3} \ \textless \ (\sqrt{6})^{-1} +(\sqrt{15})^{-1} +(\sqrt{35})^{-1} }$


[tex]\displaystyle \\ \texttt{Demonstram inegalitatea a doua} \\ \\ \frac{1}{\sqrt{6}} + \frac{1}{\sqrt{15}} + \frac{1}{\sqrt{35}} \ \textless \ \frac{16}{15} \\ \\ \frac{1}{\sqrt{6}}\ \textless \ \frac{1}{\sqrt{4}} ~~si~~ \frac{1}{\sqrt{15}}\ \textless \ \frac{1}{\sqrt{9}} ~~si~~ \frac{1}{\sqrt{35}}\ \textless \ \frac{1}{\sqrt{25}} \\ \\ \Longrightarrow ~~ \frac{1}{\sqrt{6}} + \frac{1}{\sqrt{15}} + \frac{1}{\sqrt{35}} \ \textless \ \frac{1}{\sqrt{4}} + \frac{1}{\sqrt{9}} + \frac{1}{\sqrt{25}} = [/tex]


[tex]\displaystyle \\ = \frac{1}{2} + \frac{1}{3} + \frac{1}{5}= \frac{15+10+6}{30}=\frac{31}{30}\ \textless \ \frac{32}{30}=\frac{16}{15} \\ \\ \Longrightarrow ~~ \boxed{(\sqrt{6})^{-1} +(\sqrt{15})^{-1} +(\sqrt{35})^{-1} \ \textless \ \frac{16}{15}} \\ \\ \\ a[/tex]