Solving an inequality involving floor. We have this inequality :
$$\lfloor x+1 \rfloor < \sqrt x +1.  $$
I want to find algebraic way to solve this inequality.(not geometric solution) 
Note : It is obvious when $x \in \Bbb Z$  we have $0<x<1$ but I can't solve it when $x \notin \Bbb Z$

To show another possible way , we have that
$$
\begin{array}{l}
 \left\lfloor {x + 1} \right\rfloor  < \sqrt x  + 1 \\ 
 \left\lfloor x \right\rfloor  < \sqrt x  \\ 
 x - \left\{ x \right\} < \sqrt x \quad \left| {\;0} \right. \le \left\{ x \right\} < 1 \\ 
 x - 1 < x - \left\{ x \right\} < \sqrt x  \\ 
 \end{array}
$$
where $ \left\{ x \right\}$ represents the fractional part of $x$.
For the last chain of inequalities to sussist, and since $x$ shall be non-negative, we must have:
$$
\left\{ \begin{array}{l}
 x - 1 < \sqrt x  \\ 
 0 \le x \\ 
 \end{array} \right.\quad  \Rightarrow \quad 0 \le x < 1\;\; \vee \;\;\left\{ \begin{array}{l}
 x^{\,2}  - 3x + 1 < 0 \\ 
 1 < x \\ 
 \end{array} \right.
$$
because for $0 \leqslant x<1$ the original inequality is satisfied, for $x=1$ it is not, and for $1<x$ 
we can square both sides giving the quadratic inequality.
Since the solution of the quadratic equation are
$$
\frac{{3 \pm \sqrt 5 }}{2} \approx 0.38,\;2.62
$$
and the lowest is less than $1$, we take only the higher one as the upper bound for $x$.
Because of the "squezing" $ x - 1 < x - \left\{ x \right\} < \sqrt x  $, that assures that the original 
$$
 x - \left\{ x \right\} < \sqrt x 
$$
does not have solutions outside of the range $0 < x < \frac{{3 \pm \sqrt 5 }}{2} $,
which means $ 0 \leqslant \left\lfloor x \right\rfloor \leqslant 2$.
And because
$$
\left[ {\begin{array}{*{20}c}
   {1 < x < 2} & {\left\lfloor x \right\rfloor  = 1 < \sqrt x }  \\
   {2 \le x < \frac{{3 + \sqrt 5 }}{2}} & {\left\lfloor x \right\rfloor  = 2 > \sqrt x }  \\
\end{array}} \right.
$$
we finally get that 
$$
\left\lfloor {x + 1} \right\rfloor  < \sqrt x  + 1\quad  \Rightarrow \quad 0 < x < 1\;\; \vee \;\;1 < x < 2
$$
