1
$\begingroup$

Given natural numbers $m,n,$ and a real number $a>1$, prove the inequality :

$$\displaystyle a^{\frac{2n}{m}} - 1 \geq n\big(a^{\frac{n+1}m} - a^{\frac{n-1}{m}}\big)$$

SOURCE : Inequalities (PDF) (Page Number 2 ; Question Number 153.2)

I have been trying this problem from 2 weeks but still no success. I tried every method I could think of like AM-GM, C-S, Holder and more, but could not find a proof.

Also, is it necessary for $n,m$ to be natural numbers ?

Any help will be gratefully acknowledged.

Thanks in advance ! :)

2 Answers 2

3

The answer can be given via simple calculus, and thus the result can be shown to hold true for all $x$. However, the OP has stated that he/she would prefer a solution that did not resort to calculus. So here is my edited answer. For my original answer, please check the edit history.

Let $a^{\frac{1}{m}}=x>1$. The question is equivalent to showing $$x^{2n}-1 \ge n(x^{n+1}-x^{n-1}) \iff \frac{x^{2n}-1}{x^2-1} \ge nx^{n-1}$$ Now, note that $$\frac{x^{2n}-1}{x^2-1}=\sum_{k=0}^{n-1}x^{2k}=\frac{1}{2} \left(\sum_{k=0}^{n-1}x^{2k}+x^{2n-2k-2}\right) \ge \frac{1}{2} \times 2\sum_{k=0}^{n-1}x^{n-1}=nx^{n-1}$$ From $\text{AM-GM}$. Our proof is done.

  • 1
    Isn't $g(x) = x^{2n} - nx^{n+1}+ nx^{n-1}-1$ ? That gives a different derivative. Am I overlooking something? – 2017-02-16
  • 0
    @MartinR Right, I missed a step. Does it work now? – 2017-02-16
  • 0
    @S.C.B. Any idea of a proof without calculus.... – 2017-02-16
  • 0
    @Nirbhay Do we have to do a proof without calculus? – 2017-02-16
  • 0
    @S.C.B. not really.... it is always better to show off an elegant elementary proof than a straightforward calc. answer .... Anyways your answer is very nice ... +1 :) – 2017-02-16
  • 0
    @Nirbhay lol, thanks. i will add an edit. – 2017-02-16
  • 1
    I still cannot verify your expression for $g'(x) = f''(x)$ ... – 2017-02-16
  • 0
    @MartinR Does it work now? – 2017-02-16
  • 0
    @S.C.B.: Yes... – 2017-02-16
  • 0
    @MartinR Good. That is good. – 2017-02-16
  • 0
    @Nirbhay I edited the anwer. – 2017-02-16
  • 0
    @S.C.B.: All sums should start at $k=0$. – 2017-02-16
  • 0
    @MartinR That is Done. – 2017-02-16
  • 0
    Great Answer !!! Now I am able to understand. Thanks for the help +unity – 2017-02-16
1

Let $x = a^{\frac 1m} > 1$. Using $$ x^{2n} - 1 = (x-1)(1+ x+x^2 + \ldots + x^{2n-1}) \\ x^{n+1} - x^{n-1} = (x-1) (x^{n-1}+x^n) $$ we get $$ x^{2n} - 1 - n(x^{n+1} - x^{n-1}) = (x-1)\left( 1+ x+x^2 + \ldots + x^{2n-1} - n(x^{n-1}+x^n) \right) \\ = (x-1)\sum_{k=1}^n \left( x^{k-1} + x^{2n-k} - x^{n-1}-x^n\right) \\ = (x-1)\sum_{k=1}^n (x^{n-k}-1)(x^n - x^{k-1}) \\ \ge 0 $$ (with strict inequality for $n \ge 2$), which is the desired inequality $$ x^{2n} - 1 \ge n(x^{n+1} - x^{n-1}) \, . $$ This proof works for positive real $m$ and integer $n \ge 1$. For $x < 1$ the same inequality with $\ge$ replaced by $\le$ holds.

  • 0
    Shoot, I first downvoted, and then I removed it, but there must have been an error. Sorry. – 2017-02-16