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 $I can't discern what the problem is 3. TRANSLATION: If$\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$4. Originally Posted by angel.white TRANSLATION: If$\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$EDIT: This is wrong, I misread my own translation as$\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$TRUE Because Every value of alpha leads this to be true regardless of the value of n, this statement is always true. Edit: Originally Posted by Moo *sigh* stupid double posts >.< 5. Hm, but nothing tells us that$\displaystyle \alpha$is an integer ? If you can prove that the function is increasing or decreasing, ok 6. Originally Posted by Moo Hm, but nothing tells us that$\displaystyle \alpha$is an integer ? If you can prove that the function is increasing or decreasing, ok not the best morning for me, I read it as being an element of a set containing the integers 0 and 1. Added a note stating it is not solved. 7. Listen to the unicorn, she does know all Now the problem is... understand what she says ^^ 8. Actually, your resolution can be used wisely...if the derivative (to$\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 :'(

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.