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 November 18th, 2010, 02:17 AM #1 Member   Joined: Oct 2010 Posts: 43 Thanks: 0 Nice-Please help 1. Let $f:[a,b] \to R\ (0 be a continuous function on $[a,b]$ and differentiable on $(a,b),\,f(x) <>0$ for all $x \in (a,b)$. Prove that there exists $c \in (a,b)$ so that $\frac{2}{a-c}<\frac{f'(c)}{f(c)}<\frac{2}{b-c}.$ 2. Find the least value of $A=\frac{2}{|a-b|}+\frac{2}{|b-c|}+\frac{2}{|c-a|}+\frac{5}{\sqrt{ab+bc+ca}},$ where a,b,c are real numbers satisfying $a+b+c=1$ and $ab+bc+ca>0.$
 November 18th, 2010, 02:43 AM #2 Senior Member     Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14 Re: Nice-Please help I'm really curious what does "<>" mean.
November 18th, 2010, 04:14 AM   #3
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 Originally Posted by stainburg I'm really curious what does "<>" mean.
Sorry
<> is not =.

 November 18th, 2010, 05:07 AM #4 Senior Member     Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14 Re: Nice-Please help As far as I know, I don't think there is such a function satisfied such conditions.
 November 18th, 2010, 08:23 AM #5 Global Moderator   Joined: Dec 2006 Posts: 18,250 Thanks: 1439 Why? The function f(x) = 1 satisfies them.
November 18th, 2010, 09:06 AM   #6
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Quote:
 Originally Posted by skipjack Why? The function f(x) = 1 satisfies them.
$f:[a,b] \to R\ (0, which means codomain of f(x) is the whole $\mathbb R$ , not just a single point

 November 27th, 2010, 06:43 PM #7 Global Moderator   Joined: Dec 2006 Posts: 18,250 Thanks: 1439 That's a logical interpretation, but I'm not sure whether it's the intended interpretation. Perhaps TungLHang can clarify what was meant.
November 27th, 2010, 07:15 PM   #8
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 Originally Posted by skipjack That's a logical interpretation, but I'm not sure whether it's the intended interpretation. Perhaps TungLHang can clarify what was meant.
Yeah, maybe. I compose a function g(x)=ln[f(x)], for which g'(x)=f'(x)/f(x), apply the Lagrange's mean value theorem on g(x), still find nothing. And what is the relationship between f(a) and f(b)? If not, it doesn't make any sense

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