Proof an inequality with exponentials

Question

Given $m>1$, $k\geq j \geq 0$. How to prove $\frac{m^{k+j}}{(m-1)^k(m+1)^j}\geq1$?

Edit:

If it does not hold, under what conditions of m, h, j will it hold?

Answer 1

The inequality can be rewritten as $$(m-1)^{k \over k + j}(m+1)^{j \over k + j} \leq m$$ This in turn can be rewritten as $$(m-1)^{1 \over 2}(m+1)^{1 \over 2} \bigg({m -1 \over m + 1}\bigg)^{k - j \over 2(k + j)} \leq m$$ Since $k \geq j \geq 0$ and $0 < {m - 1 \over m + 1} < 1$, the factor $\big({m -1 \over m + 1}\big)^{k - j \over 2(k + j)}$ is at most $1$, and the factor $(m-1)^{1 \over 2}(m+1)^{1 \over 2} = (m^2 - 1)^{1 \over 2}$ is less than $(m^2)^{1 \over 2} = m$. Thus the inequality holds.

Answer 2

The inequality is equivalent to

$$k\log(m-1)+j\log (m+1)\le k\log m+j\log m$$

$$k\log m-k\log(m-1) \ge j\log (m+1)-j\log m$$

$$k\log \left(\frac{m}{m-1} \right)\ge j\log \left(\frac{m+1}{m}\right)$$

wich is true since

$$\frac{m}{m-1}\ge \frac{m+1}{m} \iff m^2\ge m^2-1$$

Answer 3

hint

Put $$f(x)=\frac{x^{j+k}}{(x+1)^k(x-1)^j}$$ then

the numerator of the derivative is

$$(x+1)^{k-1}(x-1)^{j-1}x^{j+k-1}\Bigl((j+k)(x+1)(x-1)-x(k(x-1)+j(x+1))\Bigr)$$

$$=(x+1)^{k-1}(x-1)^{j-1}x^{j+k-1}\Bigl( -j-k-x(j-k) \Bigr)$$

Answer 4

Well $\frac {m^{j+k}}{(m-1)^k(m+1)^j} = (\frac m{m+1})^j* (\frac m{m-1})^k$ and so

$\frac {m^{j+k}}{(m-1)^k(m+1)^j} = (\frac m{m+1})^j* (\frac m{m-1})^k \ge 1 \iff$

$(\frac m{m-1})^k \ge (\frac {m+1}m)^j$.

And note note that $\frac m{m-1} = 1 + \frac 1{m-1}>1$ and $\frac {m+1}m = 1 +\frac 1m>1$ and that $\frac 1{m-1} > \frac 1{m}$ so as $k\ge j \ge 0$:

$(\frac m{m-1})^k=(1+\frac 1{m-1})^k \underbrace{\ge}_{\text{equality holding only if }k=j}$

$ (1+\frac 1{m-1})^j\underbrace{\ge}_{\text{equality holding only if }j=0}$

$(1 + \frac 1{m})^j = (\frac {m+1}m)^j$

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Working directly:

$\frac {m^{j+k}}{(m-1)^k(m+1)^j}= $

$(\frac m{m-1})^k(\frac {m}{m+1})^j=$

$(1+\frac 1{m-1})^k(1 -\frac 1{m+1})^j = $

$[(1+\frac 1{m-1})(1 -\frac 1{m+1})]^j*(1+\frac 1{m-1})^{k-j}\ge$

$[(1+\frac 1{m-1})(1 -\frac 1{m+1})]^j*1^{k-j} =$

$(1+\frac 1{m-1}-\frac 1{m+1} -\frac 1{(m-1)(m+1)})^j=$

$(1 + \frac {(m+1) - (m-1) - 1}{(m-1)(m+1)})^j =$

$(1 + \frac 1{(m-1)(m+1)})^j \ge 1^j = 1$.