How prove this inequality $f(x)\ge f(0)$

Question

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

Use Holder inequality and the triangle inequality? – nomen Jul 13 '14 at 03:27 — nomen, Jul 13 '14 at 03:27
@nomen,How use it? Thank you – math110 Jul 13 '14 at 03:34 — math110, Jul 13 '14 at 03:34

johannesvalks · Answer 1 · 2014-07-17T03:50:06.727

6

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

$$ f(x) \ge f(0). $$

edited Jul 17 '14 at 03:50

answered Jul 16 '14 at 00:25

johannesvalks

6,407

1

It's Nice！Thank you, +1 – math110 Jul 16 '14 at 03:28
@chinamath: You are welcome! – johannesvalks Jul 16 '14 at 12:29
1

Oh,I found this have wrong, – math110 Jul 17 '14 at 03:14
because $h_{c/b}(x)\le h_{c/b}(0)$,so $\dfrac{h_{a/b}(0)-1}{1-h_{c/b}(x)}\le \dfrac{h_{a/b}(0)-1}{1-h_{c/b}(0)}$$ – math110 Jul 17 '14 at 03:15
@chinamath: I will take a look... – johannesvalks Jul 17 '14 at 03:33
@chinamath: I have changed it. Thanks for the note... – johannesvalks Jul 17 '14 at 03:51
Hello,can you explain why $1-h_{a/b}(x)>0,1-h_{c/b}(x)>0$? – math110 Jul 17 '14 at 03:52
Let us continue this discussion in chat. – math110 Jul 17 '14 at 03:53
How do you enter the chat? – johannesvalks Jul 17 '14 at 04:07
I think this is also have wrong,because $1-h_{a/b}(x)<0$,$1-h_{c/b}>0$ – math110 Jul 17 '14 at 05:39
@chinamath: I still need to work on this issue... – johannesvalks Jul 17 '14 at 17:44
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@johannesvalks it' s fine I think. – Jul 21 '14 at 19:53
@Jung Wen Chen, I think that too - thanks for the comment! – johannesvalks Jul 21 '14 at 19:57

score 0 · Answer 2 · 2014-07-21T19:52:44.023

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 $x=1$.

How prove this inequality $f(x)\ge f(0)$

2 Answers2