Hard inequality,Please help me

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

Quote:

---

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

TRANSLATION:

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

http://www.mathhelpforum.com/math-he...nequality.html

:p

Quote:

---

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:

Quote:

---

Originally Posted by Moo

http://www.mathhelpforum.com/math-he...nequality.html

:p

---

*sigh* stupid double posts >.<

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

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

Quote:

---

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 :p

---

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.

Listen to the unicorn, she does know all :D

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

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 :'(

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 :/

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

I solved it.Now it is't hard inequality.(Clapping)

Quote:

---

Originally Posted by uyanga

I solved it.Now it is't hard inequality.(Clapping)

---

Post it, please, I'm really curious now.