How prove this inequality $f(a)\le f(b)$ Suppose $f(x)$ is continous on $[a,b]$,and for any $x_{0}\in [a,b]$.

the limit
  $$\varliminf_{x\to x^{-}_{0}}\dfrac{f(x)-f(x_{0})}{x-x_{0}}\ge 0$$
  show that
  $$f(a)\le f(b)$$

My try: I found this problem is same as
How prove this $f(a)\le f(b)$
But for my problem,this is inflimit,and the limit form is not same,Thank you 
 A: Suppose $f(b) < f(a)$. Let $m = \frac{f(b)-f(a)}{b-a} <0$.
Let $\phi(x) = f(x)+(x-a) \frac{f(a)-f(b)}{b-a}$. Note that $\phi(a)=\phi(b) = 0$. Since $\phi$ is continuous, $\phi$ has a maximum and minimum on $[a,b]$. Let $\overline{\phi} = \max \phi([a,b])$ and  $\underline{\phi} = \min \phi([a,b])$.
If $\overline{\phi} = \underline{\phi}$, then $\phi(x) = 0$ and $\liminf_{x \uparrow b} \frac{f(x)-f(b)}{x-b} = \frac{f(b)-f(a)}{b-a} = m < 0$.
If $\underline{\phi} < 0$, then let $x_0 \in (a,b)$ be a minimizer. Then $\phi(x) \ge \phi(x_0) $ for all $ x \in [a,b]$. In particular, for $x < x_0$, we have $\phi(x) - \phi(x_0) \ge 0$, and so $\liminf_{x \uparrow x_0} \frac{\phi(x)-\phi(x_0)}{x-x_0} \le 0$. Since
$f(x) -f(x_0)= \phi(x)-\phi(x_0)+(x-x_0) \frac{f(b)-f(a)}{b-a}$, we have 
$\liminf_{x \uparrow x_0} \frac{f(x)-f(x_0)}{x-x_0} \le \frac{f(b)-f(a)}{b-a} =m < 0$.
If $\overline{\phi} >0$, let $x_0 \in (a,b)$ be a maximizer. Then $\phi(x) \le \phi(x_0) $ for all $ x \in [a,b]$. In particular, $\phi(x) \le \phi(x_0)$, and so 
$f(x) -f(x_0) \le (x-x_0) \frac{f(b)-f(a)}{b-a}$, or in other words, 
$\frac{f(x)-f(x_0)}{x-x_0} \le \frac{f(b)-f(a)}{b-a} =m$ for all $x \in (x_0,b]$. Now repeat the above process on the interval $[x_0,b]$.
One of two things happens above. Either the process terminates, and we have found a point $x_0$ such that 
$\liminf_{x \uparrow x_0} \frac{f(x)-f(x_0)}{x-x_0} \le \frac{f(b)-f(a)}{b-a} =m< 0$, or the process does not terminate, and we find a sequence of points $b >x_n > x_{n-1}$ such that 
$\frac{f(x)-f(x_n)}{x-x_n} \le \frac{f(b)-f(a)}{b-a} =m$ for all $x \in (x_n,b]$.
Since $x_n$ is increasing, we have $x_n \uparrow x^* \le b$. Then the above shows that $\frac{f(x^*)-f(x_n)}{x^*-x_n} \le \frac{f(b)-f(a)}{b-a} =m$. In otherwords, 
$\liminf_{x \uparrow x^*} \frac{f(x)-f(x^*)}{x-x^*} \le \frac{f(a)-f(b)}{b-a} =m< 0$.
Consequently, if $f(b)<f(a)$, then there exists some $x_0 \in [a,b]$ such that 
$\liminf_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0} < 0$.
Hence it follows that $f(a) \le f(b)$.
