How to show that $f(x)=x+\frac{1}{x}$ is injection for $x\geq1$? I know how to start the proof, but I am running cycles and it makes me furious.
An I am also stuck on showing that this function is a surjection. 
Help would be greatly appreciated. 
 A: Suppose $y= x+ \dfrac 1 x.$ Then $x^2 -xy + 1 = 0.$ That is a quadratic equation in $x.$ Its solution for $x$ is
$$
x = \frac {y \pm \sqrt{y^2-4}} 2.
$$
If $0<y<2$ then $y^2-4<0$ and there are no real solutions. If $y>2,$ then one of the two solutions is between $0$ and $1$ and the other is $>1.$ To see that, consider what happens if you multiply the two solutions:
$$
\frac{y+\sqrt{y^2-4}} 2 \cdot \frac{y-\sqrt{y^2-4}} 2 = 1.
$$
The two solutions are reciprocals of each other; thus one is $<1$ and the other $>1.$
As far as surjectivity is concerned, notice that $x+\dfrac 1 x=2$ if $x=1,$ and if $x>1$ then
$$
x + \dfrac 1 x > x \to +\infty \text{ as } x\to+\infty,
$$
so the intermediate value theorem tells you that the image of the function is the entire interval $[2,+\infty).$
A: Suppose $x+{1\over x}=y+{1\over y}$ with $x,y\ge1$. Then $x-y={1\over y}-{1\over x}={x-y\over xy}$, hence $(xy-1)(x-y)=0$, which implies $x=y$, since $xy-1=0$ if and only if $x=y=1$ (under the given assumption that $x$ and $y$ are greater than or equal to $1$). So $x+{1\over x}$ is injective for $x\ge1$.
As for surjectivity, the question is unanswerable unless you specify the range (aka codomain).  A function is always surjective onto its image, which for $x+{1\over x}$ with domain $x\ge1$ is $[2,\infty)$, as Michael Hardy shows in his answer.
