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#1 h Dec 2, 2024, 9:48 AM • 1 Y

Y by jokehim

Problem. Let $a,b,c$ be positive real variables such that $a+b+c=3.$ Prove that $\color{black}{\sqrt{2a^{2}-a+3}+\sqrt{2b^{2}-b+3}+\sqrt{2c^{2}-c+3}\le 3\sqrt{\frac{ab+bc+ca+9}{ab+bc+ca}}.}$

This post has been edited 1 time. Last edited by KhuongTrang, Dec 29, 2024, 2:38 AM

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arqady

#2 h Dec 2, 2024, 10:14 AM

Y by

KhuongTrang wrote:

Problem. Let $a,b,c$ be real variables such that $a+b+c=3.$ Prove that $\color{black}{\sqrt{2a^{2}-a+3}+\sqrt{2b^{2}-b+3}+\sqrt{2c^{2}-c+3}\le 3\sqrt{\frac{ab+bc+ca+9}{ab+bc+ca}}.}$

The equality occurs also for $a=b=\frac{\sqrt{97}-5}{4}.$ I am not sure that there is a nice solution.

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#3 h Dec 2, 2024, 2:53 PM • 1 Y

Y by jokehim

arqady wrote:

KhuongTrang wrote:

Problem. Let $a,b,c$ be real variables such that $a+b+c=3.$ Prove that $\color{black}{\sqrt{2a^{2}-a+3}+\sqrt{2b^{2}-b+3}+\sqrt{2c^{2}-c+3}\le 3\sqrt{\frac{ab+bc+ca+9}{ab+bc+ca}}.}$

The equality occurs also for $a=b=\frac{\sqrt{97}-5}{4}.$ I am not sure that there is a nice solution.

C-S and $uvw$ help but I'd like to see something interesting :-D

:-D

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#4 h Dec 2, 2024, 4:01 PM

Y by

Easy inequality tricks :-D

:-D

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#5 h Dec 2, 2024, 4:28 PM

Y by

Very easy by C-S

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#6 h Dec 2, 2024, 6:23 PM • 1 Y

Y by jokehim

jokehim wrote:

Very easy by C-S

I know, I knew. You're the best, ok?

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arqady

#7 h Dec 2, 2024, 8:34 PM

Y by

jokehim wrote:

Very easy by C-S

If did you mean $\sum_{cyc}\sqrt{2a^2-a+3}\leq\sqrt{\sum_{cyc}(a+k)\sum_{cyc}\frac{2a^2-a+3}{a+k}},$ where $k=\frac{7+\sqrt{97}}{8},$ so it's not easy and not nice by hand, I think.

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#8 h Dec 3, 2024, 4:55 AM

Y by

arqady wrote:

jokehim wrote:

Very easy by C-S

If did you mean $\sum_{cyc}\sqrt{2a^2-a+3}\leq\sqrt{\sum_{cyc}(a+k)\sum_{cyc}\frac{2a^2-a+3}{a+k}},$ where $k=\frac{7+\sqrt{97}}{8},$ so it's not easy and not nice by hand, I think.

How is about your method uvw?

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arqady

#9 h Dec 3, 2024, 8:47 AM

Y by

$uvw$ , of course, kills it. Did you try to get a full solution?

This post has been edited 1 time. Last edited by arqady, Dec 3, 2024, 8:49 AM

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#10 h Dec 3, 2024, 12:03 PM

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arqady wrote:

$uvw$ , of course, kills it. Did you try to get a full solution?

It remains to prove $f(w^3)\ge 0$ where $f$ is a linear of $w^3.$

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arqady

#11 h Dec 3, 2024, 4:24 PM • 1 Y

Y by ehuseyinyigit

jokehim wrote:

arqady wrote:

$uvw$ , of course, kills it. Did you try to get a full solution?

It remains to prove $f(w^3)\ge 0$ where $f$ is a linear of $w^3.$

Now, I see, you did not try. :-D

:-D

This post has been edited 1 time. Last edited by arqady, Dec 3, 2024, 4:24 PM

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#12 h Dec 4, 2024, 6:19 AM • 1 Y

Y by jokehim

jokehim wrote:

arqady wrote:

$uvw$ , of course, kills it. Did you try to get a full solution?

It remains to prove $f(w^3)\ge 0$ where $f$ is a linear of $w^3.$

Are you sure?

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arqady

#13 h Dec 4, 2024, 7:37 AM

Y by

arqady wrote:

jokehim wrote:

Very easy by C-S

If did you mean $\sum_{cyc}\sqrt{2a^2-a+3}\leq\sqrt{\sum_{cyc}(a+k)\sum_{cyc}\frac{2a^2-a+3}{a+k}},$ where $k=\frac{7+\sqrt{97}}{8},$ so it's not easy and not nice by hand, I think.

It's interesting that for $b=a$ and $c=3-2a$ it remains to prove:
$(a-1)^2((2k+1)a-3k)(2ka-3-k)\geq0,$ which for $\frac{3k}{2k+1}=\frac{3+k}{2k}$ gives $k=\frac{7+\sqrt{97}}{8}$ :-D

:-D

, but the checking $c=0$ is still very ugly.

This post has been edited 1 time. Last edited by arqady, Dec 28, 2024, 2:03 PM

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arqady

#14 h Dec 28, 2024, 2:22 PM • 1 Y

Y by MyLifeMyChoice

KhuongTrang wrote:

Problem. Let $a,b,c$ be real variables such that $a+b+c=3.$ Prove that $\color{black}{\sqrt{2a^{2}-a+3}+\sqrt{2b^{2}-b+3}+\sqrt{2c^{2}-c+3}\le 3\sqrt{\frac{ab+bc+ca+9}{ab+bc+ca}}.}$

Variables should be non-negative and $ab+ac+bc\neq0$ .

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#15 h Dec 29, 2024, 2:37 AM

Y by

Also, the following inequality is also nice :-D

:-D

$\sqrt{\frac{a^2+1}{2}} + \sqrt{\frac{b^2+1}{2}} + \sqrt{\frac{c^2+1}{2}} \le \sqrt{\frac{15}{ab+bc+ca} + 4},\ \forall a,b,c>0: a+b+c=3.$

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arqady

#16 h Dec 29, 2024, 6:25 AM

Y by

KhuongTrang wrote:

Also, the following inequality is also nice :-D

:-D

$\sqrt{\frac{a^2+1}{2}} + \sqrt{\frac{b^2+1}{2}} + \sqrt{\frac{c^2+1}{2}} \le \sqrt{\frac{15}{ab+bc+ca} + 4},\ \forall a,b,c>0: a+b+c=3.$

Bacteria with $a+\frac{5+\sqrt{41}}{8}$ helps. Checking $c\rightarrow0^+$ is still very ugly.

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#17 h Dec 29, 2024, 6:49 AM

Y by

I believe there is a proof using one of Vasc's lemma.

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arqady

#18 h Dec 29, 2024, 9:21 AM • 1 Y

Y by MyLifeMyChoice

KhuongTrang wrote:

I believe there is a proof using one of Vasc's lemma.

Yes, it works! :-D

:-D

Because $g(x)=-\frac{x}{\sqrt{x^2+1}}$ is a convex function on $[0,+\infty).$

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