A cyclic inequality

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A cyclic inequality

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Source: own-CRUX

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#1 h Apr 2, 2025, 11:56 AM • 4 Y

Y by arqady, Zuyong, aidan0626, ehuseyinyigit

https://scontent.fsgn8-4.fna.fbcdn.net/v/t39.30808-6/487810123_665217679997376_7347415477366680428_n.jpg?_nc_cat=108&ccb=1-7&_nc_sid=127cfc&_nc_eui2=AeH9e18xpNQ0uUBZQQkzGI1sqihFNnXitnKqKEU2deK2crUQ9lO03mpa6uIuZrZX-fexuN7_3pgzwg_Xrjn_kJxz&_nc_ohc=KqNYPAsWn5kQ7kNvgH41EvH&_nc_oc=AdnS_8h7Z_ePKpfFxm3GPEPnomzv7IY0Al5gv7JT-KRBALVJ28_unNw2BV0v_wfzSsyCqLwvKwUvXD4QArFaY9h5&_nc_zt=23&_nc_ht=scontent.fsgn8-4.fna&_nc_gid=Wu1sAp13ECFV5g0shHydqg&oh=00_AYFfnEEVDQ4WZEG8vF9hunfc4aaA1-3EfLdz18IwZaEOVw&oe=67F309EA

This post has been edited 1 time. Last edited by KhuongTrang, Apr 2, 2025, 11:59 AM

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#2 h Apr 8, 2025, 1:22 AM

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I have five proofs. The solution submission is available untill next publication.

See some symmetrical inequalities: https://artofproblemsolving.com/community/c6h507278p2849709

This post has been edited 2 times. Last edited by KhuongTrang, Apr 8, 2025, 1:39 AM

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Quantum-Phantom

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Quantum-Phantom

#3 h Apr 8, 2025, 2:20 AM

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Can you show one or some of your proofs?

Here is my idea, 2 proofs short of completion.

We need to show that
$\sum_{\rm cyc}\sqrt{a^2+6ab}\ge\sqrt{\frac{25}2\sum_{\rm cyc}ab}=\sum_{\rm cyc}\frac{a^2+\frac{22}5ab+\frac{11}{10}ac}{\sum\limits_{\rm cyc}a^2+\frac{11}2\sum\limits_{\rm cyc}ab}\sqrt{\frac{25}2\sum_{\rm cyc}ab}.$ Let $l_a=a^2+6ab$ and
$r_a=\frac{25}2\sum_{\rm cyc}ab\left(\frac{a^2+\frac{22}5ab+\frac{11}{10}ac}{\sum\limits_{\rm cyc}a^2+\frac{11}2\sum\limits_{\rm cyc}ab}\right)^2.$ We need to show that
$\sqrt{l_a}+\sqrt{l_b}+\sqrt{l_c}\ge\sqrt{r_a}+\sqrt{r_b}+\sqrt{r_c},$ which follows from
$\begin{align*} l_a+l_b+l_c&\ge r_a+r_b+r_c,\\l_al_b+l_bl_c+l_cl_a&\ge r_ar_b+r_br_c+r_cr_a,\\l_al_bl_c&\ge r_ar_br_c. \end{align*}$ The first inequality holds, because after removing denominators we obtain
$\begin{align*} 0\le {}&\frac{1}{16}\sum _{\rm cyc}a b \left(2 a - 4 b - c\right)^{2} \left(60 a^{2} + 142 a b + 15 b^{2}\right) \\ & + \frac{1}{2}\sum_{\rm cyc} (a - b)^{2} \left( a b - 4 a c - 2 b^{2} +3 b c + 2 c^{2}\right)^{2} \\ & + \frac{1}{16}\sum_{\rm cyc}a b c \left(11804 a b^{2} + 10290 a b c + 8645 a c^{2} + 1784 c^{3}\right). \end{align*}$ However, I don't know whether the last two hold. @KhuongTrang, can you help me certify them?

This post has been edited 4 times. Last edited by Quantum-Phantom, Apr 8, 2025, 4:46 AM

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729 posts

#4 h Apr 8, 2025, 3:13 AM

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Quantum-Phantom wrote:

Can you show one or some of your proofs?

Wait for solution publication and I'll show all solutions :-D

:-D

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729 posts

#5 h Apr 8, 2025, 5:09 AM • 1 Y

Y by ehuseyinyigit

Quantum-Phantom wrote:

Can you show one or some of your proofs?

Here is my idea, 2 proofs short of completion.

We need to show that
$\sum_{\rm cyc}\sqrt{a^2+6ab}\ge\sqrt{\frac{25}2\sum_{\rm cyc}ab}=\sum_{\rm cyc}\frac{a^2+\frac{22}5ab+\frac{11}{10}ac}{\sum\limits_{\rm cyc}a^2+\frac{11}2\sum\limits_{\rm cyc}ab}\sqrt{\frac{25}2\sum_{\rm cyc}ab}.$ Let $l_a=a^2+6ab$ and
$r_a=\frac{25}2\sum_{\rm cyc}ab\left(\frac{a^2+\frac{22}5ab+\frac{11}{10}ac}{\sum\limits_{\rm cyc}a^2+\frac{11}2\sum\limits_{\rm cyc}ab}\right)^2.$ We need to show that
$\sqrt{l_a}+\sqrt{l_b}+\sqrt{l_c}\ge\sqrt{r_a}+\sqrt{r_b}+\sqrt{r_c},$ which follows from
$\begin{align*} l_a+l_b+l_c&\ge r_a+r_b+r_c,\\l_al_b+l_bl_c+l_cl_a&\ge r_ar_b+r_br_c+r_cr_a,\\l_al_bl_c&\ge r_ar_br_c. \end{align*}$ The first inequality holds, because after removing denominators we obtain
$\begin{align*} 0\le {}&\frac{1}{16}\sum _{\rm cyc}a b \left(2 a - 4 b - c\right)^{2} \left(60 a^{2} + 142 a b + 15 b^{2}\right) \\ & + \frac{1}{2}\sum_{\rm cyc} (a - b)^{2} \left( a b - 4 a c - 2 b^{2} +3 b c + 2 c^{2}\right)^{2} \\ & + \frac{1}{16}\sum_{\rm cyc}a b c \left(11804 a b^{2} + 10290 a b c + 8645 a c^{2} + 1784 c^{3}\right). \end{align*}$ However, I don't know whether the last two hold. @KhuongTrang, can you help me certify them?

I checked. Your idea works :-D

:-D

Very good application of Jichen's result.

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#6 h Apr 8, 2025, 5:16 AM

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Quantum-Phantom wrote:

We need to show that
$\sum_{\rm cyc}\sqrt{a^2+6ab}\ge\sqrt{\frac{25}2\sum_{\rm cyc}ab}=\sum_{\rm cyc}\frac{a^2+\frac{22}5ab+\frac{11}{10}ac}{\sum\limits_{\rm cyc}a^2+\frac{11}2\sum\limits_{\rm cyc}ab}\sqrt{\frac{25}2\sum_{\rm cyc}ab}.$

We just need to prove that
$\sqrt{\frac{2(a^2+6ab)}{ab+bc+ca}} \geqslant \frac{a(10a+44b+11c)}{2(a^2+b^2+c^2)+11(ab+bc+ca)}.$ The equality holds for $a=2b, \ c = 0.$

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arqady

#7 h Apr 9, 2025, 6:41 AM

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KhuongTrang wrote:

https://scontent.fsgn8-4.fna.fbcdn.net/v/t39.30808-6/487810123_665217679997376_7347415477366680428_n.jpg?_nc_cat=108&ccb=1-7&_nc_sid=127cfc&_nc_eui2=AeH9e18xpNQ0uUBZQQkzGI1sqihFNnXitnKqKEU2deK2crUQ9lO03mpa6uIuZrZX-fexuN7_3pgzwg_Xrjn_kJxz&_nc_ohc=KqNYPAsWn5kQ7kNvgH41EvH&_nc_oc=AdnS_8h7Z_ePKpfFxm3GPEPnomzv7IY0Al5gv7JT-KRBALVJ28_unNw2BV0v_wfzSsyCqLwvKwUvXD4QArFaY9h5&_nc_zt=23&_nc_ht=scontent.fsgn8-4.fna&_nc_gid=Wu1sAp13ECFV5g0shHydqg&oh=00_AYFfnEEVDQ4WZEG8vF9hunfc4aaA1-3EfLdz18IwZaEOVw&oe=67F309EA

The following inequality was in our test last year.
Let $a$ , $b$ and $c$ be non-negative numbers such that $ab+ac+bc=3$ . Prove that:
$\sqrt{a^2+24ab}+\sqrt{b^2+24bc}+\sqrt{c^2+24ca}\geq10.$ There are nice solutions for both inequalities.

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Quantum-Phantom

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Quantum-Phantom

#9 h Apr 12, 2025, 12:26 PM

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arqady wrote:

The following inequality was in our test last year.
Let $a$ , $b$ and $c$ be non-negative numbers such that $ab+ac+bc=3$ . Prove that:
$\sqrt{a^2+24ab}+\sqrt{b^2+24bc}+\sqrt{c^2+24ca}\geq10.$ There are nice solutions for both inequalities.

The method in #6 no longer works, and I do not want to try the method in #3 again. What are nice solution to these problems?

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#10 h Apr 12, 2025, 12:51 PM

Y by

Quantum-Phantom wrote:

arqady wrote:

The following inequality was in our test last year.
Let $a$ , $b$ and $c$ be non-negative numbers such that $ab+ac+bc=3$ . Prove that:
$\sqrt{a^2+24ab}+\sqrt{b^2+24bc}+\sqrt{c^2+24ca}\geq10.$ There are nice solutions for both inequalities.

The method in #6 no longer works, and I do not want to try the method in #3 again. What are nice solution to these problems?

I also want to know ,the essence of thinking to solve this problem lies.

This post has been edited 1 time. Last edited by SunnyEvan, Apr 12, 2025, 12:54 PM

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#11 h Apr 12, 2025, 1:04 PM

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Quantum-Phantom wrote:

arqady wrote:

The following inequality was in our test last year.
Let $a$ , $b$ and $c$ be non-negative numbers such that $ab+ac+bc=3$ . Prove that:
$\sqrt{a^2+24ab}+\sqrt{b^2+24bc}+\sqrt{c^2+24ca}\geq10.$ There are nice solutions for both inequalities.

The method in #6 no longer works, and I do not want to try the method in #3 again. What are nice solution to these problems?

I hope you try to prove the following inequality :-D

:-D

$\sqrt{a^2+24ab}+\sqrt{b^2+24bc}+\sqrt{c^2+24ca}\ge \sqrt{a^2+24(ab+bc+ca)}+\frac{ab+bc+ca}{a}$ where $a=\max\{a,b,c\}.$

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arqady

#12 h Apr 12, 2025, 2:18 PM • 1 Y

Y by ehuseyinyigit

Quantum-Phantom wrote:

arqady wrote:

The following inequality was in our test last year.
Let $a$ , $b$ and $c$ be non-negative numbers such that $ab+ac+bc=3$ . Prove that:
$\sqrt{a^2+24ab}+\sqrt{b^2+24bc}+\sqrt{c^2+24ca}\geq10.$ There are nice solutions for both inequalities.

The method in #6 no longer works, and I do not want to try the method in #3 again. What are nice solution to these problems?

I am ready to show. But maybe someone wants to solve this problem without a chance to see my solution. I'll show it later. OK?
Chapeau to KhuongTrang, which creates beautiful inequalities!

This post has been edited 1 time. Last edited by arqady, Apr 12, 2025, 2:24 PM

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#13 h Apr 15, 2025, 10:52 AM

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@arqady Oh, thank you! My pleasure :love:

:love:

The following inequality is also true $\sqrt{a^2+6ab+6bc}+\sqrt{b^2+6bc+6ca}+\sqrt{c^2+6ca+6ab}\ge 5+2\sqrt{3},\quad \forall a,b,c\ge 0: ab+bc+ca=2.$

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#14 h Apr 16, 2025, 1:33 PM

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KhuongTrang wrote:

https://scontent.fsgn8-4.fna.fbcdn.net/v/t39.30808-6/487810123_665217679997376_7347415477366680428_n.jpg?_nc_cat=108&ccb=1-7&_nc_sid=127cfc&_nc_eui2=AeH9e18xpNQ0uUBZQQkzGI1sqihFNnXitnKqKEU2deK2crUQ9lO03mpa6uIuZrZX-fexuN7_3pgzwg_Xrjn_kJxz&_nc_ohc=KqNYPAsWn5kQ7kNvgH41EvH&_nc_oc=AdnS_8h7Z_ePKpfFxm3GPEPnomzv7IY0Al5gv7JT-KRBALVJ28_unNw2BV0v_wfzSsyCqLwvKwUvXD4QArFaY9h5&_nc_zt=23&_nc_ht=scontent.fsgn8-4.fna&_nc_gid=Wu1sAp13ECFV5g0shHydqg&oh=00_AYFfnEEVDQ4WZEG8vF9hunfc4aaA1-3EfLdz18IwZaEOVw&oe=67F309EA

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