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#1 h Nov 20, 2009, 1:41 PM • 2 Y

Y by Adventure10, Mango247

a_i,b_i is positive real number(i=1,2...,n)
c_i=1/a_i,and C=c_1+c_2+...+c_n
d_i=1/b_i,and the same define D.
e_i=1/(a_i+b_i)and the same define E.
Prove that 1/C+1/D<=1/E.

Sorry, but I don't know how to post the topic more scienific.

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#2 h Nov 20, 2009, 1:47 PM • 1 Y

Y by Adventure10

tang zy wrote:

a_i,b_i is positive real number(i=1,2...,n)
c_i=1/a_i,and C=c_1+c_2+...+c_n
d_i=1/b_i,and the same define D.
e_i=1/(a_i+b_i)and the same define E.
Prove that 1/C+1/D<=1/E.

Sorry, but I don't know how to post the topic more scienific.

do you mean $a_i,b_i>0$ and

$\frac{1}{\sum{\frac{1}{a_i}}}+\frac{1}{\sum{\frac{1}{b_i}}} \le \frac{1}{\sum{\frac{1}{a_i+b_i}}}$

?

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#3 h Nov 20, 2009, 1:50 PM • 1 Y

Y by Adventure10

Yes,Thank you

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#4 h Nov 20, 2009, 2:40 PM • 1 Y

Y by Adventure10

$a_i,b_i > 0,i = 1,2,...,n$ , prove that
$\frac {1}{\sum_{i = 1}^n{\frac {1}{a_i}}} + \frac {1}{\sum_{i = 1}^n{\frac {1}{b_i}}} \le \frac {1}{\sum_{i = 1}^n{\frac {1}{a_i + b_i}}}.$
when $n = 1,2$ , is easy to prove that ineq holds, and we get $\frac {1}{{\frac {1}{{a_1 }} + \frac {1}{{a_2 }}}} + \frac {1}{{\frac {1}{{b_1 }} + \frac {1}{{b_2 }}}} \le \frac {1}{{\frac {1}{{a_1 + b_1 }} + \frac {1}{{a_2 + b_2 }}}}$ .

assume when $n = k$ ineq holds, i.e.:

$\frac {1}{\sum_{i = 1}^k{\frac {1}{a_i}}} + \frac {1}{\sum_{i = 1}^k{\frac {1}{b_i}}} \le \frac {1}{\sum_{i = 1}^k{\frac {1}{a_i + b_i}}},$
then, when $n = k + 1$ , we have

$\frac {1}{{\sum_{i = 1}^{k + 1} {\frac {1}{{a_i }}} }} + \frac {1}{{\sum_{i = 1}^{k + 1} {\frac {1}{{b_i }}} }}$
$= \frac {1}{{\frac {1}{{\frac {1}{{\sum_{i = 1}^k {\frac {1}{{a_i }}} }}}} + \frac {1}{{a_{k + 1} }}}} + \frac {1}{{\frac {1}{{\frac {1}{{\sum_{i = 1}^k {\frac {1}{{b_i }}} }}}} + \frac {1}{{b_{k + 1} }}}}$
$\le \frac {1}{{\frac {1}{{\frac {1}{{\sum_{i = 1}^k {\frac {1}{{a_i }}} }} + \frac {1}{{\sum_{i = 1}^k {\frac {1}{{b_i }}} }}}} + \frac {1}{{a_{k + 1} + b_{k + 1} }}}}$
$\le \frac {1}{{\frac {1}{{\frac {1}{{\sum_{i = 1}^k {\frac {1}{{a_i + b_i }}} }}}} + \frac {1}{{a_{k + 1} + b_{k + 1} }}}}$
$= \frac {1}{{\sum_{i = 1}^k {\frac {1}{{a_i + b_i }}} + \frac {1}{{a_{k + 1} + b_{k + 1} }}}}$
$= \frac {1}{{\sum_{i = 1}^{k + 1} {\frac {1}{{a_i + b_i }}} }},$
ineq aslo holds, so, by mathematical induction, ineq holds for any $n\in\mathbb{N}^+$ !!

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#5 h Nov 20, 2009, 4:21 PM • 2 Y

Y by Adventure10, Mango247

We will show that the following holds for all $i,j$ :
$\frac{a_j+b_j}{a_j b_j (a_i+b_i)} + \frac{a_i+b_i}{a_i b_i (a_j+b_j)} \leq \frac{1}{a_i b_j} + \frac{1}{a_j b_i}$
Expanding, this is
$(a_j+b_j)^2 a_i b_i + (a_i + b_i)^2 a_j b_j \leq (a_i b_j + a_j b_i)(a_i + b_i)(a_j + b_j)$ .
Several terms cancel, so:
$4 a_i a_j b_i b_j \leq (a_i b_j + a_j b_i)^2$ ,
which is true since $(a_i b_j - a_j b_i)^2$ is non-negative.

Now, taking the expression above and summing over all $i,j$ ,
$\sum_{i,j} \frac{a_j+b_j}{a_j b_j (a_i+b_i)} + \sum_{i,j} \frac{a_i+b_i}{a_i b_i (a_j+b_j)} \leq \sum_{i,j} \frac{1}{a_i b_j} + \sum_{i,j} \frac{1}{a_j b_i}$
Therefore,
$\sum_{i,j} \frac{a_j+b_j}{a_j b_j (a_i+b_i)} \leq \sum_{i,j} \frac{1}{a_i b_j}$
And so:
$(\sum_i \frac{1}{a_i}) (\sum_i \frac{1}{a_i + b_i}) + (\sum_i \frac{1}{b_i}) (\sum_i \frac{1}{a_i + b_i}) \leq (\sum_i \frac{1}{a_i}) (\sum_i \frac{1}{b_i})$ ,
which is equivalent to the stated inequality.

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#6 h Nov 20, 2009, 6:03 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

I love inequalities. Although I am now busy with school test, I still take time on trying them...

Problem. If $a_{1},a_{2},\ldots ,a_{n}$ and $b_{1},b_{2},\ldots ,b_{n}$ are positive real numbers, then
$\frac {1}{\displaystyle\underset{i = 1}{\overset{n}{\sum }}\dfrac{1}{a_{i}}} + \frac {1}{\displaystyle\underset{i = 1}{\overset{n}{\sum }}\dfrac{1}{b_{i}}}\leq \frac {1}{\displaystyle\underset{i = 1}{\overset{n}{\sum }}\dfrac{1}{a_{i} + b_{i}}}.$
Proof. Setting $\displaystyle x_{i} = \frac {1}{a_{i}}$ and $\displaystyle y_{i} = \frac {1}{b_{i}}$ for any $i = 1,2,\ldots ,n,$ the inequality becomes
$\frac {1}{\displaystyle\underset{i = 1}{\overset{n}{\sum }}x_{i}} + \frac {1}{\displaystyle\underset{i = 1}{\overset{n}{\sum }}y_{i}}\leq \frac {1}{\displaystyle\underset{i = 1}{\overset{n}{\sum }}\dfrac{x_{i}y_{i}}{x_{i} + y_{i}}},$
or
$\underset{i = 1}{\overset{n}{\sum }}\dfrac{x_{i}y_{i}}{x_{i} + y_{i}}\leq \frac {\displaystyle\left( \underset{i = 1}{\overset{n}{\sum }}x_{i}\right) \left( \underset{i = 1}{\overset{n}{\sum }}y_{i}\right) }{\displaystyle\underset{i = 1}{\overset{n}{\sum }}x_{i} + \underset{i = 1}{\overset{n}{\sum }}y_{i}}.$
This inequality can be written as
$\underset{i = 1}{\overset{n}{\sum }}\left( x_{i} - \dfrac{x_{i}y_{i}}{x_{i} + y_{i}}\right) \geq \underset{i = 1}{\overset{n}{\sum }}x_{i} - \frac {\displaystyle \left( \underset{i = 1}{\overset{n}{\sum }}x_{i}\right) \left( \underset{i = 1}{\overset{n}{\sum }}y_{i}\right) }{\displaystyle\underset{i = 1}{\overset{n}{\sum }}x_{i} + \underset{i = 1}{\overset{n}{\sum }}y_{i}},$
or
$\underset{i = 1}{\overset{n}{\sum }}\frac {x_{i}^{2}}{x_{i} + y_{i}}\geq \frac {\displaystyle\left( \underset{i = 1}{\overset{n}{\sum }}x_{i}\right) ^{2}}{\displaystyle\underset{i = 1}{\overset{n}{\sum }}x_{i} + \underset{i = 1}{\overset{n}{\sum }}y_{i}}.$
By the Cauchy-Schwarz Inequality, we have
$\underset{i = 1}{\overset{n}{\sum }}\frac {x_{i}^{2}}{x_{i} + y_{i}}\geq \frac {\displaystyle\left( \underset{i = 1}{\overset{n}{\sum }}x_{i}\right) ^{2}}{\displaystyle\underset{i = 1}{\overset{n}{\sum }}(x_{i} + y_{i})} = \frac {\displaystyle\left( \underset{i = 1}{\overset{n}{\sum }}x_{i}\right) ^{2}}{\displaystyle\underset{i = 1}{\overset{n}{\sum }}x_{i} + \underset{i = 1}{\overset{n}{\sum }}y_{i}},$
and this completes the proof.

See also here: http://canhang2007.wordpress.com/2009/11/20/inequality-74-unknown-author/

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#7 h Nov 21, 2009, 4:45 AM • 2 Y

Y by Adventure10, Mango247

very good...
learning... learning... ...

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#8 h Nov 21, 2009, 10:50 AM • 2 Y

Y by Adventure10, Mango247

You all did very well I think.
My proof is similar to can_hang's one as follow:
assume $A=\sum_{i=1}^{n}\frac{1}{a_{i}}$
$B=\sum_{i=1}^{n}\frac{1}{b_{i}}$
$C=\sum_{i=1}^{n}\frac{1}{b_{i}+a_{i}}$
then the origin
$\Leftrightarrow \frac{1}{A}+\frac{1}{B}\leq \frac{1}{C}$
$ \Leftrightarrow \(A-C)(B-C)\geq C^{2}$
$\Leftrightarrow \sum_{i=1}^{n}\frac{b_{i}}{a_{i}(a_{i}+b_{i})}\sum_{i=1}^{n}\frac{a_{i}}{b_{i}(a_{i}+b_{i})}\geq (\sum_{i=1}^{n}\frac{1}{a_{i}+b_{i}})^{2}$
By the Cauchy-Schwarz Inequality

This post has been edited 1 time. Last edited by tang zy, Nov 21, 2009, 11:48 AM

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#9 h Nov 21, 2009, 11:30 AM • 2 Y

Y by Adventure10, Mango247

Nice, and even nicer with dollar signs before and after each formula

tang zy wrote:

My proof is similar to can_hang's one as follows:
denote
$A=\sum_{i=1}^{n}\frac{1}{a_{i}}$
$B=\sum_{i=1}^{n}\frac{1}{b_{i}}$
$C=\sum_{i=1}^{n}\frac{1}{b_{i}+a_{i}}$

then the origin
$\Leftrightarrow \frac{1}{A}+\frac{1}{B}\leq \frac{1}{C}$

$\Leftrightarrow (A-C)(B-C)\geq C^{2}$

$\Leftrightarrow \sum_{i=1}^{n}\frac{b_{i}}{a_{i}(a_{i}+b_{i})}\sum_{i=1}^{n}\frac{a_{i}}{b_{i}(a_{i}+b_{i})}\geq (\sum_{i=1}^{n}\frac{1}{a_{i}+b_{i}})^{2}$

by the Cauchy-Schwarz Inequality

Note that we can iterate this inequality. E.g. putting

$A=\sum\frac{1}{a_{i}}$ , $\ B=\sum\frac{1}{b_{i}}$ , $\ C=\sum\frac{1}{c_{i}}$ , $\ S=\sum\frac{1}{a_i+b_{i}+c_{i}}$ ,

then $\frac{1}{A}+\frac{1}{B}+\frac1C\leq \frac{1}{S}$ etc.

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#10 h Nov 21, 2009, 11:53 AM • 2 Y

Y by Adventure10, Mango247

Thank you very much for your help!
I didn't use to know how to write in a scientific way

Now I can write it!!!

spanferkel wrote:

Nice, and even nicer with dollar signs before and after each formula

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#11 h Nov 24, 2009, 12:23 AM • 2 Y

Y by Adventure10, Mango247

can_hang2007 wrote:

$\underset{i = 1}{\overset{n}{\sum }}\dfrac{x_{i}y_{i}}{x_{i} + y_{i}}\leq \frac {\displaystyle\left( \underset{i = 1}{\overset{n}{\sum }}x_{i}\right) \left( \underset{i = 1}{\overset{n}{\sum }}y_{i}\right) }{\displaystyle\underset{i = 1}{\overset{n}{\sum }}x_{i} + \underset{i = 1}{\overset{n}{\sum }}y_{i}}.$

I had a different method of proving this one:

Let $X = \sum x_i, Y = \sum y_i$ . We first note that $(\frac{x_i}{X^2} + \frac{y_i}{Y^2})(\frac{1}{x_i} + \frac{1}{y_i}) \geq (\frac{1}{X} + \frac{1}{Y})^2$ for each $i$ .

Which means

$\displaystyle \frac{x_i}{X^2} + \frac{y_i}{Y^2} \geq \frac{1}{1/x_i + 1/y_i}(1/X+1/Y)^2 = \frac{x_iy_i}{x_i+y_i}(1/X+1/Y)^2$

Then, if we let $S = \frac{XY}{X+Y} \iff 1/S = 1/X+1/Y$ , then

$\displaystyle \frac{x_i}{X^2} + \frac{y_i}{Y^2} \geq \frac{x_iy_i}{(x_i+y_i)S^2}$

The above inequality holds for all $i$ , so we sum them from $i=1$ to $n$ to yield:

$\displaystyle \frac{X}{X^2} + \frac{Y}{Y^2} \geq \sum \frac{x_iy_i}{(x_i+y_i)S^2}$

But the LHS is just $1/X+1/Y = 1/S$ , so that completes the proof

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