Antiderivative of the real valued function $f$ such that $f(x)\ge {1\over x}$ Let $f:\mathbb R\rightarrow \mathbb R$  be a function satisfying $f(x)\ge {1\over x}$ for all $x\gt 0.$ Then to show that $f$ does not admit any antiderivative .
So , I thought may be assuming that it has an antiderivative  and proceeding with  that lead to some  contradiction . So, define $g(x)$ as $$g(x)=\int_c^x f(t)dt$$where $c\gt 0$  is a constant. Then $$g(x)=\int_c^x f(t)dt\\ \ge \int_c^x{dt\over t}\\=log (t)|_c^x\\=log\ x -log\ c.$$
 This does not look look any contradiction. So , can this approach at all work $?$  If so , what should I do next and if not how should I try $?$
 A: You might find Darboux's Theorem useful here: every derivative function satisfies the intermediate value property, i.e. 
$$x<y\implies\exists c\in (x,y)\text{ such that }f(c)\text{ is between }f(x)\text{ and }f(y).$$
If you can show that $f$ does not have this property (hint: consider an interval beginning at $0$), then $f$ has no antiderivative.
A: Suppose $F'(x) = f(x)$ for all $x\in \mathbb R.$ For any $x>0$ the mean value theorem shows there exists $c_x \in (0,x)$ such that
$$\frac{F(x) - F(0)}{x} = F'(c_x) = f(c_x) \ge \frac{1}{c_x} > \frac{1}{x}.$$
As $x\to 0^+,$ the left side approaces the finite value $F'(0),$ while the rightmost side approaches $\infty.$ That's a contradiction, and shows $f$ has no antiderivative on $\mathbb R.$

Previous answer, which somehow scored a $+1.$ Suppose $F'(x) = f(x)$ for all $x\in \mathbb R.$ For $x>0$ we have
$$\frac{F(x) - F(0)}{x} \ge \frac{1/x -F(0)}{x}.$$
As $x\to 0^+,$ the left side approaces the finite value $F'(0),$ while the right side, no matter what $F(0)$ equals, approaches $\infty.$ That's a contradiction, and shows $f$ has no antiderivative on $\mathbb R.$
A: I like the Darboux property applied to this.
How about the mean-value theorem too (which is at the same level
but rather more popular).  Indeed "most" of the problems about derivatives on this site can be and perhaps should be answered with a loud, cryptic "Use the MVT!"  
If there is a continuous function $F:[0,1]\to\mathbb R$ such that
$F'(x)\geq 1/x$ for all $0<x<1$ then the MVT gives
$$
\frac{F(x)-F(0)}{x-0} = F'(\xi) \geq \frac{1}\xi$$
for some $0<\xi <x$.  Consequently
$$F(x)-F(0) \geq \frac{x}{\xi} > 1$$
for all $0<x \leq 1$.  But $F(x) > F(0)+1$ is rather hard to reconcile with the fact that $F$ should be continuous.
A: As so often happens on this site the method that the OP proposed is simply
ignored in favor of our own "better" methods. 
So, perhaps, we can tolerate another answer to the problem, this time by pushing
the idea of the OP in the right direction.

Is there an antiderivative for any function $f:\mathbb R\to\mathbb R$
  for which $f(x)\geq 1/x$ for $0<x<1$?

The OP says no, and suggests a proof by contradiction. So assume otherwise, i.e., that there is an antiderivative $F$ of $f$.  

Also assume that $f$ is integrable on $[0,1]$.  See NOTE (#).

Then, for any $0<x<1$,
$$
F(1)-F(x) = \int_x^1 f(t)\,dt \geq \int_x^1 \frac1t \,dt = \log 1 - \log  x.$$
But this means that $F(x) \leq F(1) + \log x.$  We compute
$$\lim_{x\to 0+ } \log x = -\infty$$
so  $\lim_{x\to 0+ } F(x) $ is also $-\infty$ which is impossible for a continuous function.  QED

NOTE (#) But maybe not!  We assumed that $f$ is integrable on $[0,1]$.
  Is that true under these assumptions? Well $F'=f$ so $$\int f(x)\,dx =
  F(x) + C$$ so certainly $\int_a^b f(x)\,dx = F(b)-F(a).$ Isn't this
  obvious?  There are hundreds of superior mathematicians who would
  totally agree!  Unfortunately all of them died 200 years ago.  For all
   of the  18th century integration was just the reverse process of
  differentiation.  No longer. 
The Riemann integral does not integrate
  all derivatives, not even all bounded ones.  The improper Riemann
  integral does not integrate all unbounded derivatives.  Even the much
  more powerful Lebesgue integral does not integrate all unbounded
  derivatives.  So this assumption opens up a can of worms that is best
  left closed for undergraduate students.

But rather than abandon the idea, do what most of us are trained to do.  Use it as an heuristic not a proof.  We see the correct idea but have to arrive at it by a different method, in this case a simple monotonicity theorem from the calculus.
The OP's solution redux.
Assume otherwise, i.e., that there is an antiderivative $F$ of $f$.  Define $$G(x)= F(x)-\log x$$ on the interval $(0,1]$.
Then, for any $0<x<1$, $$G'(x)=F'(x)-1/x\geq 0.$$  Consequently
$G$ is nondecreasing on $(0,1]$ and hence
$$F(1)-F(x) \geq \log 1 - \log x$$
for any $0<x<1$, i.e., 
 $F(x) \leq F(1) + \log x.$  We compute
$$\lim_{x\to 0+ } \log x = -\infty$$
so  $\lim_{x\to 0+ } F(x) $ is also $-\infty$ which is impossible for a continuous function.  QED
A: Losing patience?  This is a lemon that can be squeezed just a little tiny bit more.  This answer won't get any upvotes, but some readers may be entertained.  
For most of us, asking whether $f$ has an antiderivative would usually be expressed by asking whether $f$ is a derivative.

Problem  Is there a derivative $f$ with the property that $f(x)\geq 1/x$ for all points $x$ in some interval $(0,\delta)$?

Of course not!  Not possible!  Clearly out of the question!
Just look at the associated sets
$$  \{ x: \alpha <  f(x) < \beta \} .$$
Ridiculous!  The point $0$ belongs if $\alpha<f(0)< \beta$ and is isolated on the right.
We know all about the associated sets of derivatives.  In fact, if $f$ is a derivative then the two sets
$$
\{x:\alpha <  f(x) < \beta\} \cap (x_0-\delta,x_0)
\text{ and }
\{x:\alpha <  f(x) < \beta)\} \cap (x_0,x_0+\delta)
$$
both have positive measure for all $\delta>0$ if $\alpha<f(x_0)< \beta$.  Associated sets of derivatives are quite thick at every point on both sides--even thicker than just positive measure but I'll spare you the details.
So we even know this:

Problem  Is there a derivative $f$ with the property that $f(x)\geq 1/x$ for almost every point $x$ in some interval
  $(0,\delta)$?
Answer: No.

