If $n\in N$, $n>1$ and $\alpha \in [0,1]$ prove that

$$(n^\alpha-(n-1)^\alpha)(2^\alpha+n-n^\alpha-(n-1)^\alpha-1)+(n-1)^\alpha \geq 1$$

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- Apr 12th 2008, 08:54 PMuyangaHard inequality,Please help me
If $n\in N$, $n>1$ and $\alpha \in [0,1]$ prove that

$$(n^\alpha-(n-1)^\alpha)(2^\alpha+n-n^\alpha-(n-1)^\alpha-1)+(n-1)^\alpha \geq 1$$ - Apr 12th 2008, 09:16 PMMathstud28?????????
- Apr 13th 2008, 04:00 AMangel.white
**TRANSLATION:**

If $\displaystyle n\in N$, $\displaystyle n>1$ and $\displaystyle \alpha \in [0,1]$ prove that

$\displaystyle (n^\alpha-(n-1)^\alpha)(2^\alpha+n-n^\alpha-(n-1)^\alpha-1)+(n-1)^\alpha \geq 1$ - Apr 13th 2008, 04:02 AMMoo
- Apr 13th 2008, 04:08 AMangel.white
**EDIT: This is wrong, I misread my own translation as $\displaystyle \alpha \in \{0,1\}$, this problem remains unsolved**

**Case 1. $\displaystyle \alpha = 0$**

$\displaystyle (n^0-(n-1)^0)(2^0+n-n^0-(n-1)^0-1)+(n-1)^0 \geq 1$

$\displaystyle (1-1)(1+n-1-1-1)+1 \geq 1$

$\displaystyle 1 \geq 1$

TRUE

**Case 2. $\displaystyle \alpha = 1$**

$\displaystyle (n^1-(n-1)^1)(2^1+n-n^1-(n-1)^1-1)+(n-1)^1 \geq 1$

$\displaystyle (n-(n-1))(2+n-n-(n-1)-1)+(n-1) \geq 1$

$\displaystyle (n-n+1)(2+n-n-n+1-1)+n-1 \geq 1$

$\displaystyle (1)(2-n)+n-1 \geq 1$

$\displaystyle 2-n+n-1 \geq 1$

$\displaystyle 1 \geq 1$

TRUE

Because Every value of alpha leads this to be true regardless of the value of n, this statement is always true.

**Edit:**

*sigh* stupid double posts >.< - Apr 13th 2008, 04:11 AMMoo
Hm, but nothing tells us that $\displaystyle \alpha$ is an integer ?

If you can prove that the function is increasing or decreasing, ok :p - Apr 13th 2008, 04:17 AMangel.white
- Apr 13th 2008, 04:19 AMMoo
Listen to the unicorn, she does know all :D

Now the problem is... understand what she says ^^ - Apr 13th 2008, 04:21 AMMoo
Actually, your resolution can be used wisely...if the derivative (to $\displaystyle \alpha$) of the left member is more simple to study than the function itself... dunno, i didn't manage to solve it this morning :'(

- Apr 13th 2008, 04:46 AMangel.white
I must be overthinking this, because the solution I was working on just now was much more complicated than a post in "Elementary and Middle School Math Help" should warrant :/

- Apr 13th 2008, 04:51 AMMoo
Didn't notice the category :/

Perhaps the student went into the wrong one ? - Apr 14th 2008, 06:02 PMuyangaI solved it
I solved it.Now it is't hard inequality.(Clapping)

- Apr 14th 2008, 06:19 PMangel.white