How to prove $\frac{5abc+1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1\right)+6\ge \sum_{\text{cyc}}\sqrt{5a^3+4}$ when $a^2+b^2+c^2=a+b+c$

Question

Question

Let $a,b,c>0: a^2+b^2+c^2=a+b+c.$ Prove that $$\color{black}{\frac{5abc+1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1\right)+6\ge \sqrt{5a^3+4}+\sqrt{5b^3+4}+\sqrt{5c^3+4}}.$$ I've tried to eliminate the radical by Cauchy-Schwarz$$\frac{5abc+1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1\right)+6\ge \sqrt{3}.\sqrt{5(a^3+b^3+c^3)+12}$$ But the rest is not easy to me. I am waiting some ideas and proofs.

Thanks for your help.

Updated editings: Until now, we recieved two proof by hand verified by Mathematica. Is there a simpler proof?

River Li · Answer 1 · 2023-08-06T09:21:53.937

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Update: Use a better bound.

Some thoughts.

We have, for all $x \ge 0$, $$\sqrt{5x^3 + 4} \le \frac{13}{8}x^2 - \frac34 x + \frac{17}{8}.$$ (Note: We have $\mathrm{RHS}^2 - \mathrm{LHS}^2 = \frac{169x^2 - 138x + 33}{64}(x - 1)^2 \ge 0$.)

It suffices to prove that $$\frac{5abc+1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1\right)+6 \ge \frac{13}{8}(a^2 + b^2 + c^2) - \frac34 (a + b + c) + \frac{51}{8}. \tag{1}$$ This inequality is true which can be proved by the pqr method. But my proof is not nice.

edited Aug 06 '23 at 09:21

answered Aug 05 '23 at 15:38

River Li

37,323

+1, could you please explain "Pade approximation." ? – TATA box Aug 05 '23 at 15:40
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@TATAbox You can search Pade approximation. Or see my answer here – River Li Aug 05 '23 at 15:42
Thank you very much – TATA box Aug 05 '23 at 15:47
Nice, I accept it – Dragon boy Aug 05 '23 at 17:14
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@TATAbox By the way, in your nice answer, we may alternatively use the rational bounds there (you used AM-GM). (Already upvote about two days ago.) – River Li Aug 05 '23 at 23:19
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@RiverLi It's interesting – TATA box Aug 06 '23 at 07:34
Why did you remove previous proof by Pade approximation? – TATA box Aug 06 '23 at 13:09
@TATAbox Because I found a simpler bound. You see a quadratic function bound which makes the expression much shorter. – River Li Aug 06 '23 at 13:11
Yes, I know your means but I think you should add it as alternative proof, a long with your previous one. They are helpful ideas. – TATA box Aug 06 '23 at 13:14
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@TATAbox I always want to find a better proof if any. Also, using rational bounds is well-known strategy. Anyway, thank you for the suggestion. – River Li Aug 06 '23 at 13:17
@RiverLi I have no mean seriously, just I notice that your answer is very good. It might be useful for others. – TATA box Aug 06 '23 at 13:21
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@TATAbox Thanks. I think you just take it easy. Here is a place for happy discussion. – River Li Aug 06 '23 at 13:29
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@RiverLi Of course, have an interest to share math! – TATA box Aug 06 '23 at 13:33
I really want to accept your answer @RiverLi. All answer is good TATAbox's one is so magical. – Dragon boy Aug 07 '23 at 09:45
@Dragonboy Just feel free to accept the answer you want. – River Li Aug 07 '23 at 10:44

score 1 · Accepted Answer · edited Aug 11 '23 at 04:45

Alternative proof.

By given condition, we can rewrite the OP as $$(ab+bc+ca)\left(5+\frac{1}{abc}\right)-5abc+23\ge 4\sum_{\mathrm{cyc}}\sqrt{5a^3+4}.$$ Or $$\frac{ab+bc+ca}{a+b+c}\left[5(a^2+b^2+c^2)+\frac{a^2+b^2+c^2}{abc}\right]-5abc+23\ge 4\sum_{\mathrm{cyc}}\sqrt{5a^3+4},$$ $$\iff\sum_{\mathrm{cyc}}{\left[\left(5a^2+\dfrac{a}{bc}\right)(ab+bc+ca)-5a^2bc+7a+8b+8c\right]}\ge 4(a+b+c)\sum_{\mathrm{cyc}}\sqrt{5a^3+4}.$$ Now, base on symmetrical principle we should try to prove $$\left(5a^2+\frac{a}{bc}\right)(ab+bc+ca)-5a^2bc+7a+8b+8c\ge 4(a+b+c)\sqrt{5a^3+4}. \tag{*}$$ Can you end it now ?

Indeed, the $(*)$ is equivalent to $$\left(5a^2+\dfrac{a}{bc}\right).\frac{a(b+c)}{a+b+c}+8\ge4\sqrt{5a^3+4}.$$ Notice that by AM-GM $$\frac{a(b+c)}{a+b+c}=\frac{1}{\dfrac{1}{a}+\dfrac{1}{b+c}}\ge \frac{2}{\dfrac{2}{a}+\dfrac{1}{\sqrt{bc}}}.$$ Hence, it is enough to prove$$\frac{5a^2+\dfrac{a}{bc}}{\dfrac{2}{a}+\dfrac{1}{\sqrt{bc}}}+4\ge2\sqrt{5a^3+4}.$$ Dividing both side by $a,$ we will prove $$ \frac{\dfrac{1}{bc}+5a}{\dfrac{1}{\sqrt{bc}}+\dfrac{2}{a}}+\dfrac{4}{a}\ge2\sqrt{5a+\dfrac{4}{a^2}}.$$ The last inequality is true by AM-GM$$ \frac{\dfrac{1}{bc}+5a}{\dfrac{1}{\sqrt{bc}}+\dfrac{2}{a}}+\dfrac{4}{a}= \frac{5a+\dfrac{4}{a^2}}{\dfrac{1}{\sqrt{bc}}+\dfrac{2}{a}}+\frac{1}{\sqrt{bc}}+\dfrac{2}{a}\ge2\sqrt{5a+\dfrac{4}{a^2}}.$$ The proof is done. Equality holds at $a=b=c=1.$

-1, Did you squaring both side $(*)$ ? I think it is possible because it is non homogenous inequality. — Dragon boy, Aug 06 '23 at 14:08
Please don't. I hope you can spend time to full it. Id est, I will post it. — TATA box, Aug 06 '23 at 14:15

Michael Rozenberg · Answer 3 · 2023-08-05T13:46:13.263

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A sketch of the proof.

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$ and since by AM-GM $$\sum_{cyc}\sqrt{5a^3+4}\leq\frac{1}{2}\sum_{cyc}\left(\frac{5a^3+4}{2a+1}+2a+1\right),$$ it's enough to prove that $f(w^3)\geq0,$ where $f$ decreases.

Thus, by $uvw$ it's enough to prove $f(w^3)\geq0$ for equality case of two variables, which after homogenization and assuming $b=c=1$ gives: $$(a-1)^2\left(2a^{11}+4a^{10}+11a^9-3a^8-12a^7+84a^6+464a^5+684a^4+472a^3+352a^2+32a+96\right)\geq0,$$ which is obvious.

Can you get a full solution?

edited Aug 05 '23 at 13:46

answered Aug 05 '23 at 13:37

Michael Rozenberg

194,933

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Thanks. How did you come up with such hard AM-GM ? – Dragon boy Aug 05 '23 at 13:50
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@Dragon boy Firstly I tried AM-GM with $a+2$, but it leads to a wrong inequality. But with $2a+1$ it turned out. It's interesting that C-S $\sum\limits_{cyc}\frac{a}{2a+1}\geq\frac{(a+b+c)^2}{2(a+b+c)+a+b+c}=\frac{a+b+c}{3}$ leads to a wrong inequality. – Michael Rozenberg Aug 05 '23 at 13:54
I will try expand and show it as expression of $uvw.$ The condition quite sadism – Dragon boy Aug 05 '23 at 13:57
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@Dragon boy We know that $v^2=\frac{3u^2-u}{2}$ and after this we can work with $w^3$ and $u$ only. – Michael Rozenberg Aug 05 '23 at 14:00
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@MichaelRozenberg +1, very nice AM-GM step – TATA box Aug 05 '23 at 14:26
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@TATAbox Yes, +1 for the nice AM-GM. – River Li Aug 05 '23 at 15:01
Please give me a hint https://math.stackexchange.com/questions/4748060/a-b-c-ge-0-abbcca-3-find-minimal-and-maximal-p-frac1-sqrtabc7-f – Dragon boy Aug 06 '23 at 00:15

How to prove $\frac{5abc+1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1\right)+6\ge \sum_{\text{cyc}}\sqrt{5a^3+4}$ when $a^2+b^2+c^2=a+b+c$

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