<s>[INST] 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 
 [/INST] 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
$$
f(x) \ge f(0).
$$
</s>