How prove this inequality $f(x)\ge f(0)$ Question:
let $a>b>c>0,n\in N^{+},n\ge 2$ be given numbers,show that:
$$f(x)=\dfrac{\left(\dfrac{x^a+n-1}{n}\right)^{1/a}-\left(\dfrac{x^b+n-1}{n}\right)^{1/b}}{\left(\dfrac{x^b+n-1}{n}\right)^{1/b}-\left(\dfrac{x^c+n-1}{n}\right)^{1/c}}\ge \dfrac{\left(\dfrac{n-1}{n}\right)^{1/a}-\left(\dfrac{n-1}{n}\right)^{1/b}}{\left(\dfrac{n-1}{n}\right)^{1/b}-\left(\dfrac{n-1}{n}\right)^{1/c}},x\ge 0$$
my idea : Note that
$$f(0)=\dfrac{\left(\dfrac{n-1}{n}\right)^{1/a}-\left(\dfrac{n-1}{n}\right)^{1/b}}{\left(\dfrac{n-1}{n}\right)^{1/b}-\left(\dfrac{n-1}{n}\right)^{1/c}}$$
so we must prove $f$ is non decreasing ,so we will prove 
$$f'(x)\ge 0$$
But I found $f'(x)\ge 0$ that is very ugly,(maybe someone can find a nice way to express it). Can someone help me? Thank you
@johannesvalks solution is wrong, but thank you 
 A: Given the question, let us define
$$
g_k(x) = \left( \frac{x^k + n - 1}{n} \right)^{1/k},
$$
so we can write
$$
f(x) = \frac{g_a(x) - g_b(x)}{g_b(x) - g_c(x)} =
\frac{\displaystyle \frac{g_a(x)}{g_b(x)} - 1}
{\displaystyle 1 - \frac{g_c(x)}{g_b(x)}}.
$$

Let us define
$$
h_{p/q}(x) = \frac{g_p(x)}{g_q(x)},
$$
so we obtain
$$
f(x) = \frac{h_{a/b}(x) - 1}{1 - h_{c/b}(x)}.
$$

Note that
$$
\begin{eqnarray}
h_{p/q}(x) &=& \frac{ \displaystyle \left( \frac{x^p + n - 1}{n} \right)^{1/p} }
  { \displaystyle \left( \frac{x^q + n - 1}{n} \right)^{1/q} }\\
&=& \left( \frac{x^p + n - 1}{n} \right)^{1/p}
  \left( \frac{x^q + n - 1}{n} \right)^{-1/q}\\
&=& \left( \frac{1}{n} \right)^{1/p} \left( \frac{1}{n} \right)^{-1/q}
  \Big( x^p + n - 1 \Big)^{1/p}
\Big( x^q + n - 1 \Big)^{-1/q}\\
&=& n^{1/q-1/p}\Big( x^p + n - 1 \Big)^{1/p} \Big( x^q + n - 1 \Big)^{-1/q}.
\end{eqnarray}
$$
Therefore
$$
\begin{eqnarray}
h'_{p/q}(x)
&=& n^{1/q-1/p}\Big( x^p + n - 1 \Big)^{1/p-1} \Big( x^q + n - 1 \Big)^{-1/q} x^{p-1}\\
&& \hspace{2em}
    - n^{1/q-1/p}\Big( x^p + n - 1 \Big)^{1/p} \Big( x^q + n - 1 \Big)^{-1/q-1} x^{q-1}\\
&=& h_{p/q}(x)
  \left\{ \frac{x^p}{ x^p + n - 1 } - \frac{x^q}{x^q + n - 1} \right\}\\
&=& h_{p/q}(x)
  \frac{x^p \Big( x^q + n - 1 \Big) - x^q \Big( x^p + n - 1 \Big) }
    { \Big( x^p + n - 1 \Big) \Big( x^q + n - 1 \Big) }\\
&=& h_{p/q}(x)
  \frac{ \Big( x^p - x^q \Big) \Big( n - 1 \Big) }
    { \Big( x^p + n - 1 \Big) \Big( x^q + n - 1 \Big) }.
\end{eqnarray}
$$
And as $n \ge 2$, we obtain
$$
\left[
\begin{eqnarray}
h'_{p/q}(x) \ge 0 &\textrm{for}& p>q,\\
h'_{p/q}(x) \le 0 &\textrm{for}& p<q,\\
\end{eqnarray}
\right.
$$
Whence
$$
\left[
\begin{eqnarray}
h_{p/q}(x) \ge h_{p/q}(0) &\textrm{for}& p>q,\\
h_{p/q}(x) \le h_{p/q}(0) &\textrm{for}& p<q.\\
\end{eqnarray}
\right.
$$

As $a>b>c$, we obtain
$$
\left[
\begin{eqnarray}
h_{a/b}(x) &\ge& h_{a/b}(0),\\
h_{c/b}(x) &\le& h_{c/b}(0).
\end{eqnarray}
\right.
$$

So we have
$$
\begin{eqnarray}
- f(x) &=& \frac{1 - h_{a/b}(x)}{1 - h_{c/b}(x)}\\
& \le & \frac{1 - h_{a/b}(0)}{1 - h_{c/b}(x)}\\
& \le & \frac{1 - h_{a/b}(0)}{1 - h_{c/b}(0)} = -f(0),
\end{eqnarray}
$$
whence
$$
f(x) \ge f(0).
$$
A: This is just a follow up on the previous answer.
Johannesvalks proved:
$$h_{a/b}(x)\geq h_{a/b}(0) \mbox{ and }  -h_{c/b}(x)\geq -h_{c/b}(0).$$
Suppose that we have shown that for all $x\neq1$,$x\geq0$, 
 $1< h_{a/b}(x)$ and $1>h_{c/b}(x)$. Then
$$ h_{a/b}(x) -1 \geq h_{a/b}(0)-1>0 \mbox{ and }
 1- h_{c/b}(x) \geq 1- h_{c/b}(0)>0$$
therefore
$$
\frac{h_{a/b}(x) -1}{ 1- h_{c/b}(x)} \geq \frac{h_{a/b}(0) -1}{ 1- h_{c/b}(0)},
$$
as required.
Consider as Johannesvalks
$$
g_{p}(x)=\left(\frac{x^{p}+n-1}{n}\right)^{\frac{1}{p}},
$$
We have 
$$
\partial_{p}g =  g_{p}\left(x\right)\frac{1}{p^{2}\left(x^{p}+n-1\right)}\left(x^{p}\ln\left(x^{p}\right)-\left(x^{p}+n-1\right)\ln\left(\frac{x^{p}+n-1}{n}\right)\right)$$
Now, let $t$ be given by
$$
t(y,q):=y\ln\left(y\right)-\left(y+q-1\right)\ln\left(\frac{y+q-1}{q}\right)
$$
It satisfies 
\begin{eqnarray*}
\partial_{y}t & = & \ln(y)-\ln\left(\frac{y+q-1}{q}\right)\\
\partial_{yy}t & = & \frac{q-1}{y(y+q-1)}>0
\end{eqnarray*}
As $\partial_{y}t(1,q)=0$, we have $\partial_{y}t\geq0$ for all
$y\neq1$ thus $\partial_{p}f>0$ for $x\geq0,x\neq1$. Consequently, 
$$
g_{a}(x)-g_{b}(x)>0\mbox{ and }g_{b}(x)-g_{c}(x)>0,
$$
which is what we claimed. Note that the inequality is undefined for $x=1$.
