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 , and prove that

- Apr 13th 2008, 04:02 AMMoo
- Apr 13th 2008, 04:08 AMangel.white
**EDIT: This is wrong, I misread my own translation as , this problem remains unsolved**

**Case 1.**

TRUE

**Case 2.**

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 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 ) 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