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 

For $x=0$ it is true with equality. For all $x \in (0,1]$ you have that $2-x^2>0$. Thus $$\begin{align*}\frac{x^2}{2-x^2}\le x &\iff x^2\le2x-x^3\overset{x\neq0}\iff x\le2-x^2\\& \iff x^2+x-2\le0 \\&\iff(x-1)(x+2)\le 0\phantom{X^{2^{2^{2^2}}}}\end{align*}$$ which is true since the term $(x-1)(x+2)$ changes it's sign in $x_1=-2, x_2=1$ and obviously for $x>1$ it is positive.
