Floor and fractional part functions

$\begin{aligned}\lfloor x \rfloor \times \lbrace x \rbrace &< x-1\\ \lfloor x \rfloor \times \lbrace x \rbrace &< (\lfloor x \rfloor + \lbrace x \rbrace)-1\hspace{4mm}\small\color{#3D99F6}\text{Since } x=\lfloor x \rfloor + \lbrace x \rbrace\\ 1-\lfloor x \rfloor-\lbrace x \rbrace+\lfloor x \rfloor \times \lbrace x \rbrace&<0\\ \implies (\lfloor x\rfloor-1)(\lbrace x \rbrace-1)&<0\hspace{4mm}\color{#3D99F6}\small(1)\\\\ \text{We have ,}\\ 0\leq\lbrace x \rbrace&<1\\ \implies -1\leq\lbrace x \rbrace-1&<0\hspace{4mm}\color{#3D99F6}\small(2)\\\\ \text{From }\color{#3D99F6}\small(1) \color{#333333}\normalsize\text{ and }\color{#3D99F6}\small(2)\color{#333333}\normalsize\text{ we have,}\\ (\lfloor x\rfloor-1)&>0\\ \implies\lfloor x\rfloor&>1\\ \implies x&\geq2\\ \text{Thus the minimum value for the } &\text{inequality to be satisfied occurs at }x=\color{#EC7300}\boxed{\color{#333333}2}\end{aligned}$