All of my problems on inequalities

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All of my problems on inequalities

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triangle inequality

complex numbers

inequalities unsolved

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#1 h Sep 20, 2009, 6:19 AM • 2 Y

Y by Adventure10, Mango247

problem1
originate from a friend in china,not a easy one.

$a,b,c\in R +$
$abc = 1$
proof that $\frac {1}{a} + \frac {1}{b} + \frac {1}{c} + \frac {6}{a + b + c} \geq 5$

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#2 h Sep 20, 2009, 7:08 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Start with a result of Vasc $\left(a + b\right)\left(b + c\right)\left(c + a\right) + 7\geq 5\left(a + b + c\right).$

which is equivalent with :

$\left(a + b + c\right)\left(ab + bc + ca\right) + 6\geq 5\left(a + b + c\right)\Leftrightarrow \left(ab + bc+ca \right) + \frac {6}{a + b + c}\geq 5$

done.

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#3 h Sep 20, 2009, 11:18 AM • 1 Y

Y by Adventure10

alex2008 wrote:

Start with a result of Vasc
$\left(a + b\right)\left(b + c\right)\left(c + a\right) + 7\geq 5\left(a + b + c\right).$
Click to reveal hidden text

What is Vasc's inequality?
How to proof that inequality that you write above?

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#4 h Sep 20, 2009, 1:33 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

http://www.mathlinks.ro/Forum/viewtopic.php?t=23651

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#5 h Sep 27, 2009, 4:23 PM • 2 Y

Y by Adventure10, Mango247

Pro2&3.Thanks for helping!
pro2
$\lambda \in [0,2]$
$x,y,z \in R +$
find the minnum of $\lambda$
which satisfied
$2(x^{3} + y^{3} + z^{3}) + 3xyz \geq (x^{\lambda} + y^{\lambda} + z^{\lambda})(x^{3 - \lambda} + y^{3 - \lambda} + z^{3 - \lambda})$

pro3
$\lambda , M(\lambda) , m(\lambda) \in R +$
find the mininum of $M(\lambda)$ ,and the maxinum of $m(\lambda)$
which satisfied $M(\lambda) (x + y + z)^{3} \geq x^{3} + y^{3} + z^{3} + \lambda xyz \geq m(\lambda) (x + y + z)^{3}$

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#6 h Sep 29, 2009, 9:52 AM • 2 Y

Y by Adventure10, Mango247

Any thoughts?
Please help me if you can solve

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#7 h Sep 29, 2009, 1:36 PM • 2 Y

Y by Adventure10, Mango247

Because of $abc = 1$ ,
let
$a = \frac {x^2}{yz}$ ,
$b = \frac {y^2}{zx}$ ,
$c = \frac {z^2}{xy}$ .

INE $\Leftrightarrow F(x,y,z) \ge 0$

(Now we assume that $x \le y \le z$ ,

$F(x,y,z)$
$= \sum{x^3}\sum{y^3z^3}+6x^3y^3z^3-5x^2y^2z^2\sum{x^3}$
$= F(x,x + s,x + s + t)$ (for $s,t \ge 0$ )
$= (3s^2 + 3t^2 + 3st)x^7 + (t^3 + 12st^2 + 20s^3 + 30s^2t)x^6 + (5t^4 + 62s^4 + 13st^3 + 75s^2t^2 + 124s^3t)x^5 + (270s^4t + 248s^3t^2 + 102s^2t^3 + 30t^4s + 108s^5 + 7t^5)x^4 + (109s^6 + 249s^3t^3 + 397s^4t^2 + 90t^4s^2 + 2t^6 + 20t^5s + 327s^5t)x^3 + (265s^4t^3 + 62s^7 + 125s^3t^4 + 323s^5t^2 + 217s^6t + 31s^2t^5 + 3t^6s)x^2 + (24t^5s^3 + 75s^4t^4 + 72s^7t + 3t^6s^2 + 126s^5t^3 + 126s^6t^2 + 18s^8)x + 2s^9 + 6s^4t^5 + 21s^6t^3 + t^6s^3 + 9s^8t + 18s^7t^2 + 15s^5t^4 \ge 0$

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68 posts

#8 h Sep 29, 2009, 2:26 PM • 2 Y

Y by Adventure10, Mango247

admire9898 wrote:

Pro2&3.Thanks for helping!
pro2
$\lambda \in [0,2]$
$x,y,z \in R +$
find the minnum of $\lambda$
which satisfied
$2(x^{3} + y^{3} + z^{3}) + 3xyz \geq (x^{\lambda} + y^{\lambda} + z^{\lambda})(x^{3 - \lambda} + y^{3 - \lambda} + z^{3 - \lambda})$

The inequality above is equivalent with

$x^3+y^3+z^3+3xyz\ge x^{\lambda}y^{3-\lambda}+y^{\lambda}z^{3-\lambda}+z^{\lambda}x^{3-\lambda}+y^{\lambda}x^{3-\lambda}+x^{\lambda}z^{3-\lambda}+z^{\lambda}y^{3-\lambda}$

But schur inequality gives $x^3+y^3+z^3+3xyz\ge x^2y+xy^2+yz^2+y^2z+z^2x+x^2z$

we find that the minimum of $\lambda=1$

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kuing

#9 h Sep 29, 2009, 4:15 PM • 2 Y

Y by Adventure10, Mango247

admire9898 wrote:

pro3
$\lambda , M(\lambda) , m(\lambda) \in R +$
find the mininum of $M(\lambda)$ ,and the maxinum of $m(\lambda)$
which satisfied $M(\lambda) (x + y + z)^{3} \geq x^{3} + y^{3} + z^{3} + \lambda xyz \geq m(\lambda) (x + y + z)^{3}$

use schur...

$M\left( \lambda \right)\left( {x + y + z} \right)^3 \ge x^3 + y^3 + z^3 + \lambda xyz$
$\Leftrightarrow \left( {M\left( \lambda \right) - 1} \right)\sum {x\left( {x - y} \right)\left( {x - z} \right)} + \left( {4M\left( \lambda \right) - 1} \right)\sum {\left( {y + z} \right)\left( {x - y} \right)\left( {x - z} \right)} + \left( {27M\left( \lambda \right) - 3 - \lambda } \right)xyz \ge 0$
$\Leftrightarrow \left\{ \begin{array}{l} M\left( \lambda \right) - 1 \ge 0 \\ 4M\left( \lambda \right) - 1 \ge 0 \\ 27M\left( \lambda \right) - 3 - \lambda \ge 0 \\ \end{array} \right.$
$\Leftrightarrow M\left( \lambda \right) \ge \max \left\{ {1,\frac{{3 + \lambda }}{{27}}} \right\}$
$\Rightarrow \min \left\{ {M\left( \lambda \right)} \right\} = \max \left\{ {1,\frac{{3 + \lambda }}{{27}}} \right\};$

$x^3 + y^3 + z^3 + \lambda xyz \ge m\left( \lambda \right)\left( {x + y + z} \right)^3$
$\Leftrightarrow \left( {1 - m\left( \lambda \right)} \right)\sum {x\left( {x - y} \right)\left( {x - z} \right)} + \left( {1 - 4m\left( \lambda \right)} \right)\sum {\left( {y + z} \right)\left( {x - y} \right)\left( {x - z} \right)} + \left( {3 + \lambda - 27m\left( \lambda \right)} \right)xyz \ge 0$
$\Leftrightarrow \left\{ \begin{array}{l} 1 - m\left( \lambda \right) \ge 0 \\ 1 - 4m\left( \lambda \right) \ge 0 \\ 3 + \lambda - 27m\left( \lambda \right) \ge 0 \\ \end{array} \right.$
$\Leftrightarrow m\left( \lambda \right) \le \min \left\{ {\frac{1}{4},\frac{{3 + \lambda }}{{27}}} \right\}$
$\Rightarrow \max \left\{ {m\left( \lambda \right)} \right\} = \min \left\{ {\frac{1}{4},\frac{{3 + \lambda }}{{27}}} \right\}.$

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#10 h Sep 29, 2009, 4:24 PM • 2 Y

Y by Adventure10, Mango247

matrix41 wrote:

admire9898 wrote:

Pro2&3.Thanks for helping!
pro2
$\lambda \in [0,2]$
$x,y,z \in R +$
find the minnum of $\lambda$
which satisfied
$2(x^{3} + y^{3} + z^{3}) + 3xyz \geq (x^{\lambda} + y^{\lambda} + z^{\lambda})(x^{3 - \lambda} + y^{3 - \lambda} + z^{3 - \lambda})$

The inequality above is equivalent with

$x^3 + y^3 + z^3 + 3xyz\ge x^{\lambda}y^{3 - \lambda} + y^{\lambda}z^{3 - \lambda} + z^{\lambda}x^{3 - \lambda} + y^{\lambda}x^{3 - \lambda} + x^{\lambda}z^{3 - \lambda} + z^{\lambda}y^{3 - \lambda}$

But schur inequality gives $x^3 + y^3 + z^3 + 3xyz\ge x^2y + xy^2 + yz^2 + y^2z + z^2x + x^2z$

we find that the minimum of $\lambda = 1$

how do we prove that the inequality does not hold for all $\lambda < 1$ ?

:

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130 posts

#11 h Sep 29, 2009, 5:05 PM • 2 Y

Y by Adventure10, Mango247

alex2008 wrote:

$\left(ab + bc + ca \right) + \frac {6}{a + b + c}\geq 5$
done.

How to prove the last one?

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#12 h Sep 30, 2009, 1:44 PM • 2 Y

Y by Adventure10, Mango247

4dspace wrote:

alex2008 wrote:

$\left(ab + bc + ca \right) + \frac {6}{a + b + c}\geq 5$
done.

How to prove the last one?

$\left(ab + bc + ca \right) + \frac {6}{a + b + c}\geq 5$
$\frac{ab+bc+ca}{abc} + \frac {6}{a + b + c}\geq 5$
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+ \frac {6}{a + b + c}\geq 5$

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#13 h Oct 4, 2009, 2:03 AM • 2 Y

Y by Adventure10, Mango247

Pro.4

$a,b,c\in \Re +$ Show that $(a^{2}+2)(b^{2}+2)(c^{2}+2) \geq 9(ab+bc+ca)$

Pro.5
Given $m,n,p \in Z+$ , $p\geq 3$ , $x_{1}+x_{2}+...+x_{p}=1$ , $(x_{1},x_{2},...,x_{p} \geq 0)$
find the maximum of $f(x_{1},x_{2},...,x_{p})=x^{n}_{1}x^{m}_{2}+x^{n}_{2}x^{m}_{3}+...+x^{n}_{p-1}x^{m}_{p}+x^{n}_{p}x^{m}_{1}$

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30253 posts

arqady

#14 h Oct 4, 2009, 3:42 AM • 2 Y

Y by Adventure10, Mango247

admire9898 wrote:

Pro.4

$a,b,c\in \Re +$ Show that $(a^{2} + 2)(b^{2} + 2)(c^{2} + 2) \geq 9(ab + bc + ca)$

See a stronger inequality here :
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=76508

This post has been edited 1 time. Last edited by arqady, Oct 4, 2009, 3:49 AM

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#15 h Oct 4, 2009, 3:44 AM • 2 Y

Y by Adventure10, Mango247

Pro6

$n\geq 2$ , $n \in Z$ . $x_{1},x_{2},...,x_{n}\in Z+$ .
$x_{1} \leq x_{2} \leq ... x_{n}$ .And
$x_{1}+x_{2}+...+x_{n}=x_{1}x_{2}...x_{n}$ .
find the maximum value of $x_{1}+x_{2}...+x_{n}$

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30253 posts

arqady

#16 h Oct 4, 2009, 3:50 AM • 2 Y

Y by Adventure10, Mango247

4dspace wrote:

alex2008 wrote:

$\left(ab + bc + ca \right) + \frac {6}{a + b + c}\geq 5$
done.

How to prove the last one?

Let $f(ab + ac + bc) = ab + bc + ca + \frac {6}{a + b + c}.$ Hence, $f$ is an increasing function.
Thus, $f$ gets a minimal value, when $ab + ac + bc$ gets a minimal value, id est,
when two numbers from $\{a,b,c\}$ are equal.
$ab + bc + ca + \frac {6}{a + b + c}\geq 5\Leftrightarrow\frac {1}{a} + \frac {1}{b} + \frac {1}{c} + \frac {6}{a + b + c}\geq\frac {5}{\sqrt [3]{abc}}.$
Let assume $b = c = 1$ in the last inequality.
Id est, it remains to prove that $\frac {1}{a} + 2 + \frac {6}{a + 2}\geq\frac {5}{\sqrt [3]a},$ which is true.

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#17 h Oct 4, 2009, 7:52 AM • 2 Y

Y by Adventure10, Mango247

Pro.7

Try to show that

│x+z│+│y+z│+│x+y│ $\leq$ │x│+│y│+│z│+│x+y+z│

$x,y,z \in C$

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#18 h Oct 4, 2009, 8:52 AM • 2 Y

Y by Adventure10, Mango247

admire9898 wrote:

Pro.7

Try to show that

│x+z│+│y+z│+│x+y│ $\leq$ │x│+│y│+│z│+│x+y+z│

$x,y,z \in C$

does Triangle Inequality hold for complex numbers?

if so, then square both sides and we are done.

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#19 h Oct 4, 2009, 10:02 AM • 2 Y

Y by Adventure10, Mango247

earldbest wrote:

admire9898 wrote:

Pro.7

Try to show that

│x+z│+│y+z│+│x+y│ $\leq$ │x│+│y│+│z│+│x+y+z│

$x,y,z \in C$

does Triangle Inequality hold for complex numbers?

if so, then square both sides and we are done.

Yes. As a matter of fact it was posted here:

http://www.mathlinks.ro/viewtopic.php?search_id=2092480903&t=303580

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#20 h Oct 13, 2009, 12:18 PM • 2 Y

Y by Adventure10, Mango247

Pro8

Attachments:

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#21 h Oct 13, 2009, 12:55 PM • 2 Y

Y by Adventure10, Mango247

$ab+bc+ca+a+b+c\ge6\sqrt{abc}=6=3abc+3=3(abc+1)$

$(ab+bc+ca+a+b+c)+3(ab+bc+ca+a+b+c)\ge3(a+b+c+ab+bc+ca)+3(abc+1)$

$4(ab+bc+ca+a+b+c)\ge3(ab+bc+ca+a+b+c+abc+1)=3(a+1)(b+1)(c+1)$

$4(a(c+1)+b(a+1)+c(b+1))\ge3(a+1)(b+1)(c+1)$

$\frac{a(c+1)+b(a+1)+c(b+1)}{(a+1)(b+1)(c+1)}\ge\frac34$

$\frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)}\ge\frac34$

clear

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#22 h Oct 13, 2009, 2:29 PM • 2 Y

Y by Adventure10, Mango247

this is equivalent to $4\left(\sum a + \sum \frac {1}{a}\right) \ge 3\left(2 + \sum a + \sum \frac {1}{a}\right)$ , which is obvious by AM-GM

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#23 h Oct 18, 2009, 2:43 AM • 2 Y

Y by Adventure10, Mango247

Thanks!
Good proof!

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