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 
 A: All you really need to show is that $f(x) \le x$ in $[0,1)$, since the rest is trivial, $f(x) \ge 0\ \forall\ x: |x|<\sqrt2$ and since $x<1$ in $[0,1)$ you only need the stronger bound.
To see $f(x) \le x$ (with strict inequality in the interior $(0,1)$), look at
$$\frac{x^2}{2-x^2} \le_0 \frac{x^2}{2- 1^2} =  x^2 \le_0 x$$
Where $\le_0$ denotes $<$ on $(0,1)$ and $=$ for $x=0$.
A: $$0\leqslant x<1\Rightarrow 0\leqslant x^2<x<1\\2-x^2>1\\\frac{1}{2-x^2}<1\\\text{multiply by }x^2\\x^2\frac{1}{2-x^2}<x^2*1\\\frac{x^2}{2-x^2}<x^2\leqslant x <1\\$$
A: 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.
A: 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 $(0,1)$.
