amparvardi wrote: Let $x > 1$ be a non-integer number. Prove that \[\biggl( \frac{x+\{x\}}{[x]} - \frac{[x]}{x+\{x\}} \biggr) + \biggl( \frac{x+[x]}{ \{x \} } - \frac{ \{ x \}}{x+[x]} \biggr) > \frac 92 \] I found a better constant: \[\biggl( \frac{x+\{x\}}{\lfloor x\rfloor} - \frac{\lfloor x\rfloor}{x+\{x\}} \biggr) + \biggl( \frac{x+\lfloor x\rfloor}{ \{x \} } - \frac{ \{ x \}}{x+\lfloor x\rfloor} \biggr) > \frac{14}{3} \] We have: \[LHS= 2\left(\frac{\{x\}}{\lfloor x\rfloor}+\frac{\lfloor x\rfloor}{\{x\}}\right)+2\left(\frac{\{x\}}{2\{x\}+\lfloor x\rfloor}+\frac{\lfloor x\rfloor}{2\lfloor x\rfloor+\{x\}}\right)> 4+2\left(\frac{\{x\}}{3\lfloor x\rfloor}+\frac{\lfloor x\rfloor}{3\lfloor x\rfloor}\right)=4+\frac{2x}{3\lfloor x\rfloor}\ge 4+\frac{2}{3}=\frac{14}{3}\] I used AM-GM, $x=\lfloor x\rfloor+\{x\}$, $\lfloor x\rfloor>\{x\}$ and $x\ge \lfloor x\rfloor$, for $x>1$

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=525986&p=2980692#p2980692 Let $x > 1$ be a non-integer number. Prove that\[\biggl( \frac{x+\{x\}}{[x]} - \frac{[x]}{x+\{x\}} \biggr) + \biggl( \frac{x+[x]}{ \{x \} } - \frac{ \{ x \}}{x+[x]} \biggr) > \frac{16}{3} \]

ionbursuc wrote: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=525986&p=2980692#p2980692 Let $x > 1$ be a non-integer number. Prove that\[\biggl( \frac{x+\{x\}}{[x]} - \frac{[x]}{x+\{x\}} \biggr) + \biggl( \frac{x+[x]}{ \{x \} } - \frac{ \{ x \}}{x+[x]} \biggr) > 5 \]

Crux M292: Let $x > 0$ . Prove that\[\sqrt{ \frac{[x]}{x+\{x\}}}+ \sqrt{ \frac{\{x\}}{x+[x]} } > 1. \]

Let $ a$={$x$},$b$=math=inline]$x$[/math,we have $a>0$ ,$b\ge 1$ you ineq is $ \frac{b+2a}{b}+\frac{2b+a}{a}-\frac{b}{b+2a}-\frac{a}{2b+a}>\frac{9}{2} $ or $ 2(\frac{a}{b}+\frac{b}{a})-\frac{b}{b+2a}-\frac{a}{2b+a}>\frac{5}{2} $ but we know $\frac{b}{b+2a}+\frac{a}{2b+a}<1 $ so we have $ 2(\frac{a}{b}+\frac{b}{a})-\frac{b}{b+2a}-\frac{a}{2b+a}>3 $ so $ \frac{b+2a}{b}+\frac{2b+a}{a}-\frac{b}{b+2a}-\frac{a}{2b+a}>5 $

$\clubsuit \color{green}{\textit{\textbf{Proof:}}}$ Since \[\frac{x+\{x\}}{\lfloor x \rfloor}+\frac{x+\lfloor x\rfloor}{\{x\}}=1+\frac{2\{x\}}{\lfloor x\rfloor}+\frac{2\lfloor x\rfloor}{\{x\}}+1 \ge 1+4+1=6,\]we now show that \[\frac{\lfloor x \rfloor}{x+\{x\}}+\frac{\{x\}}{x+\lfloor x\rfloor}< \frac{3}{2}.\]Since $x>1$, we have \[\frac{\lfloor x \rfloor}{x+\{x\}}+\frac{\{x\}}{x+\lfloor x\rfloor}<\frac{x}{x}+\frac{1}{x+1}<\frac{3}{2}. \quad \blacksquare\]