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 

Since $f(0)=0$ we can reduce ourselves to consider $f$ restricted to $(0,1]$.
Now $0\leq f(x) \leq x$ in $(0,1)$ if and olny if $0\leq f(x)/{x} \leq 1$ in $(0,1)$.
So we consider the function $$g(x)=\frac{f(x)}{x}.$$
As you showed $g$ is increasing in $(0,1]$ and one sees that $g(1)=1$. This shows that $0\leq g(x) \leq 1$ in $(0,1)$.
