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A long time ago, I saw it in a AOPS forum. I've found the original link but it is no longer viewable.

Question. For all $a,b,c>0: a+b+c=ab+bc+ca$. Prove that $$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}+33\ge 4\left(\sqrt{\frac{5}{a}+4}+\sqrt{\frac{5}{b}+4}+\sqrt{\frac{5}{c}+4}\right)$$

I want to find some elegant proofs for this nice problem. It would be great if there's an AM-GM or Cauchy-Schwarz ideas.

Firstly, I denote $a+b+c=p=ab+bc+ca=q; abc=r.$

By using Cauchy-Schwarz inequality, we'll prove$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}+33\ge 4\sqrt{3}\sqrt{\frac{5}{a}+\frac{5}{b}+\frac{5}{c}+12}.$$Or$$\color{black}{33+\frac{q}{r}(q-2r)=33+\frac{q^2-2pr}{r}\ge 4\sqrt{\frac{15q}{r}+36}.}$$ It's easy to see that $p=q\ge 3.$ By squaring both side, it turns out $$513+\left(\frac{q^2-2qr}{r}\right)^2+66\frac{q^2-2qr}{r}-\frac{240q}{r}\ge 0.$$I stopped here since the last inequality is quite complicated.

Also, I tried the old result $$\color{black}{\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge\sqrt{3(a^2+b^2+c^2)}}.$$ See MSE and more AOPS.

Now, it's enough to prove $$\sqrt{3(a^2+b^2+c^2)}+33\ge 4\left(\sqrt{\frac{5}{a}+4}+\sqrt{\frac{5}{b}+4}+\sqrt{\frac{5}{c}+4}\right).$$ I was stuck here.

Recalled by Michael Rozenberg's proofs, I think the following will help.

By AM-GM, we may find an estimate $$2\sqrt{\frac{5}{a}+4}=2\sqrt{\frac{\dfrac{5}{a}+4}{xa+y}\cdot(xa+y)}\le \frac{\dfrac{5}{a}+4}{xa+y}+(xa+y).$$ We'll choose $x+y=3.$ The rest can be solved by $pqr.$

In cases there's exist $(x,y),$ I hope the solver will give your motivation to obtain that approach.

That's all what I've done so far. Thank you for your contributions.

Blue
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Dragon boy
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3 Answers3

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I have a partial result in the case $a,b,c\ge 1.$ One of them has to be since their sum is greater equal $3$, but I haven't found an argument for all of them.

We get from Maclaurin's inequality $$ \dfrac{a+b+c}{3} \ge \sqrt{\dfrac{ab+bc+ca}{3}}=\sqrt{\dfrac{a+b+c}{3}} $$ and therefore $a+b+c\ge 3.$ If $a,b,c\ge 1$ then $$ \sqrt{\dfrac{5}{x}+4} \le \sqrt{5+4}=3 \;\text{ for all }\;x\in \{a,b,c\} $$ and again by Maclaurin $$ \begin{align*} \dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ca}{b}+33&\ge \sqrt{3}\cdot \sqrt{a^2+b^2+c^2}+33\\ &=\sqrt{3}\cdot\sqrt{(a+b+c)^2-2(a+b+c)}+33\\ &=\sqrt{3}\cdot\sqrt{(a+b+c-1)^2-1^2}+33\\ &=\sqrt{3}\cdot\sqrt{\left[(a+b+c)\cdot \underbrace{(a+b+c-2)}_{\ge 1}\right]}+33 \\ &\geq \sqrt{3}\cdot\sqrt{a+b+c}+33\\ &\ge 36 \\ &=4\cdot (3+3+3)\\ &\ge 4\cdot \left(\sqrt{\dfrac{5}{a}+4}+\sqrt{\dfrac{5}{b}+4}+\sqrt{\dfrac{5}{c}+4}\right) \end{align*} $$

As mentioned below, we cannot achieve $a,b,c\ge 1$ except for the trivial case $a=b=c=1.$ However, the statement still follows if we can prove $$ 9\ge \sqrt{\dfrac{5}{a}+4}+\sqrt{\dfrac{5}{b}+4}+\sqrt{\dfrac{5}{c}+4} $$ which at least simplifies the expression. Assume $a\ge b\ge c>0.$

Newton's inequality states $$ \dfrac{(ab+bc+ca)^2}{9}\ge abc\cdot \dfrac{a+b+c}{3}=abc \cdot\dfrac{ab+bc+ca}{3} $$ so $ab+bc+ca=a+b+c\ge 3abc$ and $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge 3.$ Since $a+b+c\ge 3$ we have $a\ge 1$ and $\dfrac{1}{a}\le 1.$ \begin{align*} 1+\dfrac{1}{b}+\dfrac{1}{c}&\ge \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge 3\\ \dfrac{2}{c}&=\dfrac{1}{c}+\dfrac{1}{c}\ge \dfrac{1}{b}+\dfrac{1}{c}\ge 2 \\ \dfrac{1}{c} &\ge 1 \\ \end{align*} Thus $a\ge 1 \ge c >0.$

Marius S.L.
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  • $a, b, c \ge 1$ and $a + b + c = ab + bc + ca$ will imply $a = b = c = 1$ (easy; e.g. let $a = 1 + s, b = 1 + t, c = 1 + r$ for $s, t, r \ge 0$, then from $a + b + c = ab + bc + ca$, we have $r + s + t + rs + st + tr = 0$ which results in $r = t = s = 0$). So actually you are just proving that the inequality is true when $a = b = c = 1$. – River Li Dec 10 '23 at 23:04
  • @River Li: Well, yes and no. If we drop the assumption that all three are greater than one, we still have the calculation that reduces the statement to $9\ge \sum_{cyc}\sqrt{\dfrac{5}{a}+4}$. That's at least more than what we started from since we are left with three separate variables. Then it can be shown by Newton's inequality - assumed $a\ge b\ge c >0$ - that $a\ge 1\ge c$ and $b$ is somewhere in between. I have shown that we have to be more careful with the margins. Only the comfortable way that all three are nice doesn't work. – Marius S.L. Dec 10 '23 at 23:47
  • I think that you need to include the result for $a \ge 1 \ge c$ and $b$ between, or add some explanation about your purpose of this answer. Because in the current version, "I have an partial answer in the case $a, b, c \ge 1$" is a trivial result (equivalently $a = b = c = 1$). – River Li Dec 10 '23 at 23:58
  • @RiverLi I added the argumentation. – Marius S.L. Dec 11 '23 at 00:15
  • You said "However, the statement still follows if we can prove $9\ge \sqrt{\dfrac{5}{a}+4}+\sqrt{\dfrac{5}{b}+4}+\sqrt{\dfrac{5}{c}+4}$". How about $a = 2, b = 2 - \frac{1}{10000}, c = \frac{1}{29999}$? In other words, $\min(a,b,c)$ can be very near zero. – River Li Dec 11 '23 at 00:27
  • Does that fulfill the condition $a+b+b=ab+bc+ca$ which I extensively used in my calculations? – Marius S.L. Dec 11 '23 at 00:32
  • It fulfills the condition $a + b + c = ab + bc + ca$ exactly. – River Li Dec 11 '23 at 00:46
  • In which case the problem is less symmetric than I thought, i.e. I covered not all solutions. I said it was a partial result. If your example is one - means: it is too late here to check it right now - then it can be used to show where the symmetry breaks in Maclaurin's and Newton's formulas. I assume at $a+b+c - 2\sim 1.$ There are further options at that step. – Marius S.L. Dec 11 '23 at 00:57
  • Unfortunately :) If we use Computer Algebra System (CAS), we know that $9> \sqrt{\dfrac{5}{a}+4}+\sqrt{\dfrac{5}{b}+4}+\sqrt{\dfrac{5}{c}+4}$ never occurs. It holds that: if $a, b, c > 0$ and $a + b + c = ab + bc + ca$, then $\sqrt{\dfrac{5}{a}+4}+\sqrt{\dfrac{5}{b}+4}+\sqrt{\dfrac{5}{c}+4} \ge 9$. – River Li Dec 11 '23 at 01:15
  • So much to an elegant solution. Another approach is to take an idea of the proof of those inequalities and consider the polynomial $(x-a)(x-b)(x-c)$ and examine its behavior. – Marius S.L. Dec 11 '23 at 01:37
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    Perhaps it works but I am afraid that it is complicated. For example, we can eliminate the root sign by $\frac{5}{a] + 4 = (2 + x)^2$ etc. However, it is complicated. – River Li Dec 11 '23 at 03:00
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Relevant proof. See my answer in similar answer.

It is also an isolated fudging as RiverLi's answer.

Proof.

By multiplying $ab+bc+ca$ and replace the hypothesis, we'll prove$$\sum_{cyc}\left[\left(\frac{bc}{a}+5\right)(b+c)+8(ab+bc+ca)\right]\ge 4(ab+bc+ca)\sum_{cyc}\sqrt{\frac{5}{a}+4}.$$Now, it's enough to show$$\left(\dfrac{bc}{a}+5\right).\frac{b+c}{ab+bc+ca}+8\ge 4\sqrt{\frac{5}{a}+4}. \tag{*}$$

Apply AM-GM inequality$$\frac{b+c}{ab+bc+ca}=\frac{1}{a+\dfrac{bc}{b+c}}\ge \frac{1}{a+\dfrac{bc}{2\sqrt{bc}}}=\frac{2}{2a+\sqrt{bc}}.$$ It suffices to prove$$\frac{\dfrac{bc}{a}+5}{\sqrt{bc}+2a}+4\ge 2\sqrt{\frac{5}{a}+4}$$ or $$\iff\frac{bc+5a}{\sqrt{bc}+2a}+4a\ge 2\sqrt{5a+4a^2}$$$$\iff \sqrt{bc}+2a+\frac{5a+4a^2}{\sqrt{bc}+2a}\ge 2\sqrt{5a+4a^2},$$which is obvious by AM-GM.

Thus, the proof is done. Equality holds iff $a=b=c=1.$

TATA box
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Some thoughts.

We use the so-called isolated fudging.

After homogenization, it suffices to prove that, for all $a, b, c > 0$, $$\left(\sum_{\mathrm{cyc}}\frac{ab}{c} \right)\frac{a + b + c}{ab + bc + ca} + 33 \ge \sum_{\mathrm{cyc}} 4\sqrt{\left(\frac{5}{a} + \frac{4(a + b + c)}{ab + bc + ca}\right)\frac{ab + bc + ca}{a + b + c}}$$ or \begin{align*} &\sum_{\mathrm{cyc}} \left(\frac{3a^2(b^3 + c^3) - 14a^2bc(b + c) + 34ab^2c^2 + 3b^2c^2(b+c)}{6abc(ab + bc + ca)} + 11\right)\\[6pt] \ge{}& \sum_{\mathrm{cyc}} 4\sqrt{\left(\frac{5}{a} + \frac{4(a + b + c)}{ab + bc + ca}\right)\frac{ab + bc + ca}{a + b + c}}. \end{align*}

It suffices to prove that, for all $a, b, c > 0$, \begin{align*} & \frac{3a^2(b^3 + c^3) - 14a^2bc(b + c) + 34ab^2c^2 + 3b^2c^2(b+c)}{6abc(ab + bc + ca)} + 11\\[6pt] \ge{}& 4\sqrt{\left(\frac{5}{a} + \frac{4(a + b + c)}{ab + bc + ca}\right)\frac{ab + bc + ca}{a + b + c}}. \tag{1} \end{align*} (1) is true which is verified by Mathematica.

River Li
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