Some cyclic inequalities

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Some cyclic inequalities

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inequalities proposed

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#1 h Sep 8, 2010, 6:25 PM • 4 Y

Y by Adventure10, Sgg88, and 2 other users

Let $a,$ $b,$ $c$ be positive real numbers. Prove that

(a) $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{13}{9}(a+b+c) \ge \frac{22}{3}\cdot \frac{a^2+b^2+c^2}{a+b+c};$

(b) $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{13}{5}\cdot \frac{ab+bc+ca}{a^2+b^2+c^2} \ge \frac{28}{5};$

(c) $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+1 \ge 4\sqrt{\frac{a^2+b^2+c^2}{ab+bc+ca}};$

(d) $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \ge \frac{3(a^3+b^3+c^3)}{a^2+b^2+c^2};$

(e) $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \ge 3\sqrt{\frac{a^4+b^4+c^4}{a^2+b^2+c^2}}.$

Please try to prove these inequalities. After you have found the solutions, please try also to find CS solutions to them (notice that for each inequality, there exists at least one CS solution).

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arqady

#2 h Sep 9, 2010, 2:25 AM • 2 Y

Y by Adventure10, Mango247

can_hang2007 wrote:

Let $a,$ $b,$ $c$ be positive real numbers. Prove that

(a) $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{13}{9}(a+b+c) \ge \frac{22}{3}\cdot \frac{a^2+b^2+c^2}{a+b+c};$

$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge \frac{37(a^2+b^2+c^2)-19(ab+ac+bc)}{6(a+b+c)}$ is stronger and still true.

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#3 h Sep 9, 2010, 11:19 AM • 1 Y

Y by Adventure10

arqady wrote:

can_hang2007 wrote:

Let $a,$ $b,$ $c$ be positive real numbers. Prove that

(a) $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{13}{9}(a+b+c) \ge \frac{22}{3}\cdot \frac{a^2+b^2+c^2}{a+b+c};$

$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge \frac{37(a^2+b^2+c^2)-19(ab+ac+bc)}{6(a+b+c)}$ is stronger and still true.

Have you nice proof, dear arqady? Thanks.

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arqady

#4 h Sep 9, 2010, 1:12 PM • 2 Y

Y by Adventure10, Mango247

can_hang2007 wrote:

arqady wrote:

$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge \frac{37(a^2+b^2+c^2)-19(ab+ac+bc)}{6(a+b+c)}$ is stronger and still true.

Have you nice proof, dear arqady? Thanks.

My proof is not nice. Have you a CS's proof of this inequality, dear Can?

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#5 h Sep 9, 2010, 1:17 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

arqady wrote:

can_hang2007 wrote:

arqady wrote:

$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge \frac{37(a^2+b^2+c^2)-19(ab+ac+bc)}{6(a+b+c)}$ is stronger and still true.

Have you nice proof, dear arqady? Thanks.

My proof is not nice. Have you a CS's proof of this inequality, dear Can?

Sorry, I have only CS proof for
$\sum \frac{a^2}{b} +k(a+b+c) \ge \frac{3(k+1)(a^2+b^2+c^2)}{a+b+c},$
where $k \le \frac{1+2\sqrt{6}}{4} \approx 1.47474\ldots$

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abch42

#6 h Sep 10, 2010, 9:44 AM • 2 Y

Y by Adventure10, Mango247

Another inequality for the collection
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\frac{(a+b+c)\left(a^2+b^2+c^2\right)}{ab+bc+ca}$ .

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abch42

#7 h Sep 10, 2010, 10:17 AM • 1 Y

Y by Adventure10

Let $x=\sqrt{\frac{a^2+b^2+c^2}{ab+bc+ca}}$ . Prove that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge x^2+x+1$ .

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#8 h Sep 10, 2010, 10:22 AM • 1 Y

Y by Adventure10

my dear arqady , those of them are non-symmetry

Can you post your answer?

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#9 h Sep 10, 2010, 12:59 PM • 2 Y

Y by Adventure10, Mango247

abch42 wrote:

Let $x=\sqrt{\frac{a^2+b^2+c^2}{ab+bc+ca}}$ . Prove that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge x^2+x+1$ .

Why not
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{a^2+b^2+c^2}{ab+bc+ca}+2\sqrt{\frac{a^2+b^2+c^2}{ab+bc+ca}}$

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#10 h Sep 10, 2010, 1:14 PM • 2 Y

Y by Adventure10, Mango247

Tourish wrote:

abch42 wrote:

Let $x=\sqrt{\frac{a^2+b^2+c^2}{ab+bc+ca}}$ . Prove that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge x^2+x+1$ .

Why not
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{a^2+b^2+c^2}{ab+bc+ca}+2\sqrt{\frac{a^2+b^2+c^2}{ab+bc+ca}}$

It can be proved by CS.

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#11 h Sep 10, 2010, 1:35 PM • 2 Y

Y by Adventure10, Mango247

Congratulations!In fact,my solution to these inequalites is very ugly....

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375 posts

#12 h Sep 17, 2010, 2:17 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

can_hang2007 wrote:

Let $a,$ $b,$ $c$ be positive real numbers. Prove that

(d) $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \ge \frac{3(a^3+b^3+c^3)}{a^2+b^2+c^2};$

Please try to prove these inequalities. After you have found the solutions, please try also to find CS solutions to them (notice that for each inequality, there exists at least one CS solution).

My solution using CS :

Attachments:

Problem1.pdf (70kb)

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#13 h Sep 17, 2010, 2:28 PM • 2 Y

Y by Adventure10, Mango247

minhhoang wrote:

can_hang2007 wrote:

Let $a,$ $b,$ $c$ be positive real numbers. Prove that

(d) $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \ge \frac{3(a^3+b^3+c^3)}{a^2+b^2+c^2};$

Please try to prove these inequalities. After you have found the solutions, please try also to find CS solutions to them (notice that for each inequality, there exists at least one CS solution).

My solution using CS :

Very good solution, Minhhoang. It's different and nicer than mine. Congrats.

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#14 h Sep 27, 2010, 12:56 PM • 3 Y

Y by luofangxiang, Adventure10, and 1 other user

Tourish wrote:

Why not
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{a^2+b^2+c^2}{ab+bc+ca}+2\sqrt{\frac{a^2+b^2+c^2}{ab+bc+ca}}$

Let me post now the solution to this inequality.

Multiplying each side of the inequality by $ab+bc+ca$ and noting that
$\frac{a(ab+bc+ca)}{b}=\frac{ca^2}{b}+ca+a^2,$
we can rewrite it as
$\frac{ab^2}{c}+\frac{bc^2}{a}+\frac{ca^2}{b}+ab+bc+ca \ge 2\sqrt{(a^2+b^2+c^2)(ab+bc+ca)}.$
Now, substituting $a:=\frac{1}{a},$ $b:=\frac{1}{b}$ and $c:=\frac{1}{c},$ it becomes
$\frac{c}{ab^2}+\frac{a}{bc^2}+\frac{b}{ca^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca} \ge 2\sqrt{\frac{(a^2b^2+b^2c^2+c^2a^2)(a+b+c)}{a^3b^3c^3}},$
or
$a^3b+b^3c+c^3a+abc(a+b+c) \ge 2\sqrt{abc(a^2b^2+b^2c^2+c^2a^2)(a+b+c)}.$
The last one rewrites as
$ab(c^2+a^2)+bc(a^2+b^2)+ca(b^2+c^2) \ge 2\sqrt{abc(a^2b^2+b^2c^2+c^2a^2)(a+b+c)}.$
Setting $A=ab,$ $B=bc,$ $C=ca,$ $y=c^2,$ $z=a^2$ and $x=b^2,$ the above inequality becomes
$A(y+z)+B(z+x)+C(x+y) \ge 2\sqrt{(AB+BC+CA)(xy+yz+zx)},$ (1)
which is the known Ukrainian Mathematical Olympiad 2001.

And I said the problem can be proved by CS because (1) can be proved by CS. Also, there exists another CS solution but it is longer.

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#15 h Sep 27, 2010, 4:09 PM • 2 Y

Y by Adventure10, Mango247

can_hang2007 wrote:

Let $a,$ $b,$ $c$ be positive real numbers. Prove that

(e) $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \ge 3\sqrt{\frac{a^4+b^4+c^4}{a^2+b^2+c^2}}.$

Please try to prove these inequalities. After you have found the solutions, please try also to find CS solutions to them (notice that for each inequality, there exists at least one CS solution).

This ineq can be prove by the way which similar to # 12 , but it's longer, too.

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30252 posts

arqady

#16 h Aug 13, 2012, 8:34 AM • 2 Y

Y by Adventure10, Mango247

can_hang2007 wrote:

Let $a,$ $b,$ $c$ be positive real numbers. Prove that
(e) $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \ge 3\sqrt{\frac{a^4+b^4+c^4}{a^2+b^2+c^2}}.$

See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2739178#p2739178
It's ugly, but it's proof.

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#17 h Dec 16, 2016, 3:24 PM • 3 Y

Y by adityaguharoy, Adventure10, Mango247

can_hang2007 wrote:

Tourish wrote:

Why not
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{a^2+b^2+c^2}{ab+bc+ca}+2\sqrt{\frac{a^2+b^2+c^2}{ab+bc+ca}}$

Let me post now the solution to this inequality.

Multiplying each side of the inequality by $ab+bc+ca$ and noting that
$\frac{a(ab+bc+ca)}{b}=\frac{ca^2}{b}+ca+a^2,$
we can rewrite it as
$\frac{ab^2}{c}+\frac{bc^2}{a}+\frac{ca^2}{b}+ab+bc+ca \ge 2\sqrt{(a^2+b^2+c^2)(ab+bc+ca)}.$
Now, substituting $a:=\frac{1}{a},$ $b:=\frac{1}{b}$ and $c:=\frac{1}{c},$ it becomes
$\frac{c}{ab^2}+\frac{a}{bc^2}+\frac{b}{ca^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca} \ge 2\sqrt{\frac{(a^2b^2+b^2c^2+c^2a^2)(a+b+c)}{a^3b^3c^3}},$
or
$a^3b+b^3c+c^3a+abc(a+b+c) \ge 2\sqrt{abc(a^2b^2+b^2c^2+c^2a^2)(a+b+c)}.$
The last one rewrites as
$ab(c^2+a^2)+bc(a^2+b^2)+ca(b^2+c^2) \ge 2\sqrt{abc(a^2b^2+b^2c^2+c^2a^2)(a+b+c)}.$
Setting $A=ab,$ $B=bc,$ $C=ca,$ $y=c^2,$ $z=a^2$ and $x=b^2,$ the above inequality becomes
$A(y+z)+B(z+x)+C(x+y) \ge 2\sqrt{(AB+BC+CA)(xy+yz+zx)},$ (1)
which is the known Ukrainian Mathematical Olympiad 2001.

And I said the problem can be proved by CS because (1) can be proved by CS. Also, there exists another CS solution but it is longer.

//cdn.artofproblemsolving.com/images/1/e/5/1e57b2c3a591c52eba130c4ccfd3c005ed8f0206.jpg

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#18 h Dec 16, 2016, 3:31 PM • 3 Y

Y by luofangxiang, Adventure10, Mango247

Very nice luofangxiang.

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xzlbq

#19 h Dec 16, 2016, 11:57 PM • 2 Y

Y by Adventure10, Mango247

can_hang2007 wrote:

Tourish wrote:

abch42 wrote:

Let $x=\sqrt{\frac{a^2+b^2+c^2}{ab+bc+ca}}$ . Prove that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge x^2+x+1$ .

Why not
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{a^2+b^2+c^2}{ab+bc+ca}+2\sqrt{\frac{a^2+b^2+c^2}{ab+bc+ca}}$

It can be proved by CS.

best is:

${\frac {x}{y}}+{\frac {y}{z}}+{\frac {z}{x}}\geq \sqrt {4\,{\frac {{x}^{2} +{y}^{2}+{z}^{2}}{xy+zx+yz}}+2-2\,{\frac {xy+zx+yz}{{x}^{2}+{y}^{2}+{z }^{2}}}}+{\frac {{x}^{2}+{y}^{2}+{z}^{2}}{xy+zx+yz}}$

This post has been edited 2 times. Last edited by xzlbq, Dec 16, 2016, 11:57 PM

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