Study of a function on interval $[0,1]$ Let $f(x)$ be a function defined on the interval $[0,1]$ such that 
$$x \mapsto \dfrac{x^2}{2-x^2}$$
Show that, for all  $x \in [0,1[,\ 0\leq f(x)\leq x <1.$
Attempt:
Let $$g(x)=\dfrac{f(x)}{x} $$
$$g'(x)=\dfrac{2+x^2}{(2-x^{2})^2}\geq 0 $$ 
i'm stuck here 
Thanks for your help 

$$0\leqslant x<1\Rightarrow 0\leqslant x^2<x<1\\2-x^2>1\\\frac{1}{2-x^2}<1\\\text{multiply by }x^2\\x^2\frac{1}{2-x^2}<x^2*1\\\frac{x^2}{2-x^2}<x^2\leqslant x <1\\$$
