Minimum Value of a Non-symmetric Expression

freemind

#1

Minimum Value of a Non-symmetric Expression

**Posted:** Sat Mar 29, 2008 4:45 pm

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mehdi cherif

#2

$\frac {4\sqrt {2}}{n}$ ??? if it's true i write my sollution Wink

**Posted:** Sat Mar 29, 2008 5:33 pm

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andyciup

#3

Nice little problem Smile

We shall prove that the minimum value of the expression is $n\cdot\frac{\sqrt{17}}{2}$ , with equality iff all the numbers are equal.

By the inequality $x_{1}^{2}+x_{2}^{2}+\cdots+x_{m}^{2}\geq\frac{(x_{1}+x_{2}+\cdots+x_{m})^{2}}{m}$ , applied for $m=17$ , $x_{1}=a_{1}$ , and $x_{2}=x_{3}=\dots=x_{17}=\frac{1}{4a_{2}}$ , we obtain $\displaystylea_{1}^{2}+16\cdot\frac{1}{16a_{2}^{2}}\geq\frac{(a_{1}+\frac{4}{a_{2}})^{2}}{17},$ hence we have obtained

$\sqrt{a_{1}^{2}+\frac{1}{a_{2}^{2}}}\geq\frac{1}{\sqrt{17}}\cdot\left(a_{1}+\frac{4}{a_{2}}\right)$ . Similarily we have:

$\sqrt{a_{i}^{2}+\frac{1}{a_{i+1}^{2}}}\geq \frac{1}{\sqrt{17}}\cdot\left(a_{i}+\frac{4}{a_{i+1}}\right)$

and thus by summing all these relations we obtain $LHS\geq\frac{1}{\sqrt{17}}\left(a_{1}+a_{2}+\cdots+a_{n}+\frac{4}{a_{1}}+\frac{4}{a_{2}}+\cdots+\frac{4}{a_{n}}\right)$ $(\star)$ .

But $a_{i}+\frac{1}{4a_{1}}\geq 1$ , for all $i$ , and since $\frac{15}{4}\cdot\left(\frac{1}{a_1}+\cdots+\frac{1}{a_n}\right)\geq\frac{15n^{2}}{4(a_{1}+\cdots+a_{n})}\geq\frac{15n}{2},$ by summing we obtain

$a_{1}+a_{2}+\cdots+a_{n}+\frac{4}{a_{1}}+\frac{4}{a_{2}}+\cdots+\frac{4}{a_{n}}\geq n\cdot\frac{17}{2}$ , therefore, from the relation $(\star)$ , we have $LHS\geq n\cdot\frac{\sqrt{17}}{2}$ , which is indeed what we wanted to prove.

This is indeed the minimum, as the equality occurs for all the numbers equal to $\frac{1}{2}$ .

**Posted:** Sat Mar 29, 2008 5:50 pm

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NguyenDungTN

#4

Re: Minimum Value of a Non-symmetric Expression

**Posted:** Sat Mar 29, 2008 6:44 pm

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w00t?

Nguyen Manh Dung Cool (speak)

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silouan

#5

Re: Minimum Value of a Non-symmetric Expression

**Posted:** Sat Mar 29, 2008 8:11 pm

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dduclam

#6

Re: Minimum Value of a Non-symmetric Expression

**Posted:** Sat Mar 29, 2008 10:13 pm

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Duong Duc Lam

Inequalities Master · Yang-Mills Theory Offline Joined: 16 Feb 2008 Posts: 572

#7

I think it is a very weak problem for a TST.
Because it is very easy to prove that:
$\dfrac{1}{a_1}+...+\dfrac{1}{a_n}\geq2n$ and
$\sqrt{a_1^2+\dfrac{1}{a_2^2}}\geq\dfrac{(a_1+\dfrac{4}{a_2})}{\sqrt{17}}$

**Posted:** Mon Mar 31, 2008 3:46 pm