$f(x)= \frac{\sqrt{x^2-1}}{x+\log x}$ in the set $E=[1,+ \infty)$ I have the function $f(x)= \frac{\sqrt{x^2-1}}{x+\log x}$ in the set $E=[1,+ \infty)$
(i) I have to prove that $\forall y_0 \in [0,1) $exists only one $x_0 \in E $ such that $f(x_0)=y_0$
(ii) to discuss the uniform continuity of f in E.
Supposing that there are two different values  $x_1$ and $x_2 \in E $ such that 
$f(x_1)=y_0=f(x_2)$, I have to prove that $   f(x_1)=y_0=f(x_2)>1$
Can someone help me to understand what to do?

(i) you can show $f'(x)>0 \forall x \in E$:
$$f'(x)=\frac{-x^2+x^2ln(x)+x+1}{x\sqrt{x^2-1}(x+ln(x))^2}>0\Leftrightarrow-x^2+x^2ln(x)+x+1>0\Leftrightarrow ln(x)+\frac{1}{x}+\frac{1}{x^2}>1$$
Now let $g(x)=ln(x)+\frac{1}{x}+\frac{1}{x^2}$. You can show that g(x) has a local minimum at $x=2$:
$$g'(x)=0\Leftrightarrow \frac{1}{x}-\frac{1}{x^2}-\frac{2}{x^3}=0\Leftrightarrow\frac{x^2-x-2}{x^3}=0\Leftrightarrow (x-2)(x+1)=0\Leftrightarrow x=2$$
$$g''(2)=\frac{3}{8}>0$$
$$\Rightarrow g(x)\geq g(2)=\frac{3}{4}+ln(2)>1$$
$\Rightarrow f(x)$ is injective
