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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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
If , and prove that
Originally Posted by angel.white TRANSLATION:
If , and prove that EDIT: This is wrong, I misread my own translation as , this problem remains unsolved Case 1.
TRUE Case 2.
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 http://www.mathhelpforum.com/math-he...nequality.html *sigh* stupid double posts >.<
Last edited by angel.white; Apr 13th 2008 at 04:20 AM.
Hm, but nothing tells us that is an integer ?
If you can prove that the function is increasing or decreasing, ok
Originally Posted by Moo Hm, but nothing tells us that 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.
Listen to the unicorn, she does know all
Now the problem is... understand what she says ^^
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 :'(
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.
Originally Posted by uyanga I solved it.Now it is't hard inequality. Post it, please, I'm really curious now.
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