Does the implication $x>1\implies x>1/x$ need to be proven? Does this seemingly evident fact,

for $x > 1, (x > (1/x))$  

need to proven?
 A: If you can't prove it immediately, then yes, it needs to be proven. Saying something is "self-evident" to get around having to prove it is just going to get you into trouble.
A: A simple proof:
Note $x>1\Rightarrow 1=\frac{x}{x}>\frac{1}{x}$. Putting it all together, we conclude $x>1>\frac{1}{x}$
In general, to prove this kind of statements, a "clever" multiplication/division on both sides is, sometimes, a useful trick.
A: $$ \text{suppose      } x > 1 \implies   x \leq \frac{1}{x}$$
$$ x \leq \frac{1}{x}$$
$$ \implies x^{2} \leq 1 $$
$$ \implies x \leq 1 $$
$$ →← $$
$$ \text{so,  } x > 1 \implies x > \frac{1}{x} $$
A: Follows from that fact that f(x)=(1/x) is decreasing for x>0. Graphically:

A: Some of the most "obvious" facts are the most frustrating to prove (possibly because one is only asked to prove them when one has little to no experience with proofs, in the first place)! Thus, it is always better to assume that something needs to be proved, unless one already has the result from a previous proof.
Assuming that $x>1,$ we can prove each of the following from first principles (and, of course, use earlier points together with first principles to prove subsequent points):


*

*$x>0$

*$x^2>1$

*$\frac1x>0$

*$x>\frac1x$


Added: There are four more fairly crucial facts needed to prove that first point: $0x=0,$ $-1\cdot x=-x,$ $-(-x)=x,$ and $0<1.$ The last is often taken as an axiom, but not always. (Typically, it is only assumed that $0\neq 1,$ since otherwise all real numbers are equal.) Proof of the first is fairly standard: $$0=-(0x)+0x=-0x+(0+0)x=-(0x)+0x+0x=0+0x=0x.$$ Then (skipping some steps), we have $$-1\cdot x=-1\cdot x+x+-x=-1x+1x+-x=(-1+1)x+-x=0x+-x=-x.$$ Next (again skipping some steps), $$-(-x)=-(-x)+-x+x=-1(-x)+1(-x)+x=(-1+1)(-x)+x=0(-x)+x=x.$$ Finally, since the reals are totally ordered, we then must have $0<1$ or $1<0.$ But if $1<0,$ then $$0=1+-1<0+-1=-1,$$ so $$0=0\cdot-1<-1\cdot-1=-(-1)=1,$$ so $0<0,$ a contradiction.
I leave the rest to you.
A: Past a certain stage of "mathematical maturity", of course you don't need to prove it – but you should prove it now. (See various comments and answers: if you have to ask, then yes.) For your own sake as well, so that you can know it's true and why, rather than assume it. 
There are many ways to establish this, but among the most straightforward is by appealing to:
$$
\text{if } 0 < x < y \text{ then } 0 < \frac 1 y < \frac 1 x
$$
So if you have:
$$
0 < 1 < x
$$
you can conclude:
$$
0 < \frac 1 x < \frac 1 1 = 1
$$
