If $n\in N$, $n>1$ and $\alpha \in [0,1]$ prove that
$$(n^\alpha-(n-1)^\alpha)(2^\alpha+n-n^\alpha-(n-1)^\alpha-1)+(n-1)^\alpha \geq 1$$

2. ## ?????????

Originally Posted by uyanga
If $n\in N$, $n>1$ and $\alpha \in [0,1]$ prove that
$$(n^\alpha-(n-1)^\alpha)(2^\alpha+n-n^\alpha-(n-1)^\alpha-1)+(n-1)^\alpha \geq 1$$
With all these \$ I can't discern what the problem is

3. TRANSLATION:

If $n\in N$, $n>1$ and $\alpha \in [0,1]$ prove that
$(n^\alpha-(n-1)^\alpha)(2^\alpha+n-n^\alpha-(n-1)^\alpha-1)+(n-1)^\alpha \geq 1$

4. Originally Posted by angel.white
TRANSLATION:

If $n\in N$, $n>1$ and $\alpha \in [0,1]$ prove that
$(n^\alpha-(n-1)^\alpha)(2^\alpha+n-n^\alpha-(n-1)^\alpha-1)+(n-1)^\alpha \geq 1$
EDIT: This is wrong, I misread my own translation as $\alpha \in \{0,1\}$, this problem remains unsolved

Case 1. $\alpha = 0$
$(n^0-(n-1)^0)(2^0+n-n^0-(n-1)^0-1)+(n-1)^0 \geq 1$

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

$1 \geq 1$

TRUE

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

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

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

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

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

$1 \geq 1$

TRUE

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

Edit:
Originally Posted by Moo
*sigh* stupid double posts >.<

5. Hm, but nothing tells us that $\alpha$ is an integer ?

If you can prove that the function is increasing or decreasing, ok

6. Originally Posted by Moo
Hm, but nothing tells us that $\alpha$ is an integer ?

If you can prove that the function is increasing or decreasing, ok
not the best morning for me, I read it as being an element of a set containing the integers 0 and 1.

Added a note stating it is not solved.

7. Listen to the unicorn, she does know all

Now the problem is... understand what she says ^^

8. Actually, your resolution can be used wisely...if the derivative (to $\alpha$) of the left member is more simple to study than the function itself... dunno, i didn't manage to solve it this morning :'(

9. 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 :/

10. Didn't notice the category :/
Perhaps the student went into the wrong one ?

11. ## I solved it

I solved it.Now it is't hard inequality.

12. Originally Posted by uyanga
I solved it.Now it is't hard inequality.
Post it, please, I'm really curious now.