# Inequality $\sum\limits_{cyc}\frac{a^3}{13a^2+5b^2}\geq\frac{a+b+c}{18}$

Let $$a$$, $$b$$ and $$c$$ be positive numbers. Prove that: $$\frac{a^3}{13a^2+5b^2}+\frac{b^3}{13b^2+5c^2}+\frac{c^3}{13c^2+5a^2}\geq\frac{a+b+c}{18}$$

This inequality is strengthening of the following Vasile Cirtoaje's one, which he created in 2005.

Let $$a$$, $$b$$ and $$c$$ be positive numbers. Prove that: $$\frac{a^3}{2a^2+b^2}+\frac{b^3}{2b^2+c^2}+\frac{c^3}{2c^2+a^2}\geq\frac{a+b+c}{3}.$$

My proof of this inequality you can see here: https://artofproblemsolving.com/community/c6h22937p427220

But this way does not help for the starting inequality.

A big problem we have around the point $$(a,b,c)=(0.785, 1.25, 1.861)$$ because the difference between the LHS and the RHS in this point is $$0.0000158...$$.

I tried also to use Cauchy-Schwarz, but without success.

Also, I think the BW (see here https://math.stackexchange.com/tags/buffalo-way/info I tryed!) does not help.

• Is the function $f : (a,b,c) \rightarrow \sum\limits_{cyc} \frac{6a^3}{13a^2+5b^2}$ concave? Also, C-S = "Cauchy-Schwarz inequality"? Avoid this kind of unconventional abbreviation, please. – Vincent May 8 '16 at 18:17
• @Santropedro There is the following known inequality $\sum\limits_{cyc}\frac{a^3}{a^2+b^2}\geq\frac{a+b+c}{2}$. I think, it's just interesting, for which maximal value of $k$ the following inequality is still true. $\sum\limits_{cyc}\frac{a^3}{ka^2+b^2}\geq\frac{a+b+c}{k+1}$. – Michael Rozenberg Apr 19 '17 at 17:28
• @Santropedro The value $\,k=2.6\,$ in the OP is pretty sharp. I found that $k=2.603279$ is already beyond: Then Michael's given $(0.785, 1.25, 1.861)$ violates the ineq.. – Hanno May 11 '17 at 18:24
• Further result obtained via random-based simulation: $(0.785, 1.2535, 1.873)$ doesn't satisfy the inequality if $\,k=2.603262\,$. Below that $k$-value I didn't detect a violation. – Hanno May 11 '17 at 18:50
• @max8128 I don't like it. I am sorry. – Michael Rozenberg May 24 '17 at 10:19

@Maxim Gilula asked a good question: What is the point of presenting a solution via computer?

Let me address it from two aspects.

First, from my experience (the problems from MSE, AoPS, academic research etc.), the inequalities with elegant solutions (e.g. by hand) are usually those designed e.g. for contest. Except for the designed inequalities, the inequalities require usually the help of computer. Many inequalities in MSE, AoPS are open for many years without any solution by hand.

Second, although also with the help of computer, the solutions are different.

1) Some solutions e.g. Buffalo Way (BW) or Sum of Squares (SOS) provide step-by-step rigorous complete analytical solutions with detailed explanation. Moreover, these solutions are often not very long. Usually one or several A4 pages are enough to write down the step-by-step rigorous complete analytical solutions with detailed explanation.

2) In contrast, some solutions does not provided step-by-step solutions. For example, the step-by-step solutions are in Mathematica's innards, and for some cases, Mathematica run several hours to output 'true' which means that the step-by-step solutions may be very, very long, probably 10,000 A4 pages are required to write down the step-by-step solutions. Sometimes the method of Lagrange Multiplier works well, however, sometimes it results in very complicated equations which requires Mathematica etc.

3) Often, although the thinking process is very complicated and requires the help of computer, the solution is elementary and easy to check even by hand. For example, recently, I used computer to solve an inequality in AoPS to get the identity $$a^3 + b^3 + c^3 - a^2b - b^2c - c^2a = \frac{(a^2+b^2-2c^2)^2 + 3(a^2-b^2)^2 + \sum_{\mathrm{cyc}} 4(a+b)(c+a)(a-b)^2}{8(a+b+c)}$$. In contrast, Mathematica only outputs 'true'. Which solution is better?

EDIT

Remark: Actually, the Buffalo Way works though the solution is ugly. I do not put it here. With computer, here is an SOS (Sum of Squares) solution:

WLOG, assume that $$c = \min(a, b, c)$$. We have \begin{align*} &\frac{a^3}{13a^2+5b^2} + \frac{b^3}{13b^2+5c^2} + \frac{c^3}{13c^2+5a^2} - \frac{a+b+c}{18}\\ =\ & \frac{1}{18(13a^2+5b^2)(13b^2+5c^2)(13c^2+5a^2)}\\ &\quad \cdot \frac{1}{2223}\Big(cz_1^TA_1 z_1 + c(a-c)(b-c)z_2^TA_2z_2 + (b-c)z_3^TA_3z_3 + (a-c)z_4^TA_4z_4\Big) \end{align*} (Remark: $$A_1, A_2, A_3, A_4$$ are all positive definite. So the inequality is true.) where $$z_1 = \left(\begin{array}{c} {\left(a - c\right)}^2\, \left(b - c\right)\\ \left(a - c\right)\, {\left(b - c\right)}^2\\ c\, {\left(a - c\right)}^2\\ c\, \left(a - c\right)\, \left(b - c\right)\\ c\, {\left(b - c\right)}^2\\ c^2\, \left(b - c\right)\\ c^2\, \left(a - c\right) \end{array}\right), \quad z_2 = \left(\begin{array}{c} {\left(a - c\right)}^2\\ \left(a - c\right)\, \left(b - c\right)\\ {\left(b - c\right)}^2\\ c\, \left(a - c\right)\\ c^2\\ c\, \left(b - c\right) \end{array}\right),$$ $$z_3 = \left(\begin{array}{c} {\left(a - c\right)}^2\, \left(b - c\right)\\ \left(a - c\right)\, {\left(b - c\right)}^2\\ c\, {\left(a - c\right)}^2\\ c\, \left(a - c\right)\, \left(b - c\right)\\ c\, {\left(b - c\right)}^2\\ c^2\, \left(b - c\right)\\ c^2\, \left(a - c\right) \end{array}\right) , \quad z_4 = \left(\begin{array}{c} {\left(a - c\right)}^2\, \left(b - c\right)\\ \left(a - c\right)\, {\left(b - c\right)}^2\\ c\, {\left(a - c\right)}^2\\ c\, \left(a - c\right)\, \left(b - c\right)\\ c\, {\left(b - c\right)}^2\\ c^2\, \left(b - c\right)\\ c^2\, \left(a - c\right) \end{array}\right),$$ and $$A_1 = \left(\begin{array}{ccccccc} 453245 & -100035 & 111397 & 113607 & -146718 & 24687 & 6498\\ -100035 & 166231 & -19773 & 24453 & -82004 & -11286 & 31356\\ 111397 & -19773 & 444600 & 86526 & -144261 & 13585 & -6669\\ 113607 & 24453 & 86526 & 760500 & -344736 & -126711 & 149188\\ -146718 & -82004 & -144261 & -344736 & 297882 & -26676 & -53352\\ 24687 & -11286 & 13585 & -126711 & -26676 & 160056 & -111150\\ 6498 & 31356 & -6669 & 149188 & -53352 & -111150 & 160056 \end{array}\right),$$ $$A_2 = \left(\begin{array}{cccccc} 281580 & 33098 & -64467 & 40261 & 4275 & -24219\\ 33098 & 329004 & -124982 & 13572 & 20241 & -51129\\ -64467 & -124982 & 106704 & -51129 & -6840 & -30875\\ 40261 & 13572 & -51129 & 326610 & -8645 & -79794\\ 4275 & 20241 & -6840 & -8645 & 62244 & -46683\\ -24219 & -51129 & -30875 & -79794 & -46683 & 153216 \end{array}\right) ,$$ $$A_3 = \left(\begin{array}{ccccccc} 258609 & -100035 & 102011 & 75582 & -56069 & -57798 & 126945\\ -100035 & 55575 & -37791 & -29393 & 2223 & 26923 & -26676\\ 102011 & -37791 & 589342 & 224757 & -195624 & -155376 & 313272\\ 75582 & -29393 & 224757 & 693576 & -279851 & -320283 & 535977\\ -56069 & 2223 & -195624 & -279851 & 200070 & 91143 & -306945\\ -57798 & 26923 & -155376 & -320283 & 91143 & 293436 & -435708\\ 126945 & -26676 & 313272 & 535977 & -306945 & -435708 & 1016158 \end{array}\right) ,$$ $$A_4 = \left(\begin{array}{ccccccc} 144495 & -57057 & 3705 & 24206 & -22230 & 2457 & 26923\\ -57057 & 55575 & 4199 & -15561 & -22724 & 13338 & -1989\\ 3705 & 4199 & 200070 & 18031 & -46449 & -9063 & 17784\\ 24206 & -15561 & 18031 & 351468 & -140049 & -6435 & -13509\\ -22230 & -22724 & -46449 & -140049 & 189202 & -117990 & -46449\\ 2457 & 13338 & -9063 & -6435 & -117990 & 177840 & -48659\\ 26923 & -1989 & 17784 & -13509 & -46449 & -48659 & 253422 \end{array}\right).$$

• If you are allowed to use a computer, there are 100 different solutions. What is the point of presenting a solution via computer? Might as well copy and paste wolfram alpha "minimize f(x,y,z)." This doesn't deserve 100 bounty -- no answer here does. Lagrange may be less work and can be actually done by hand if you somehow memorized how to solve quartic equations and doesn't involve a computer, or once you're at that step use a computer. – Maxim Gilula Mar 18 at 20:17
• @MaximGilula Thank you for comments. Let me reply in the beginning of the solution because the comments are too long. – River Li Mar 19 at 1:03
• Thanks for your useful remarks! – Maxim Gilula Mar 19 at 1:17
• @MaximGilula Welcome to discuss. This inequality is waiting for elegant solutions (e.g. by hand) for many years. – River Li Mar 19 at 1:23
• I have found this link mathworld.wolfram.com/WitchofAgnesi.html . Where we recognize (the cartesian equation) an element of the LHS.So do you think there exists a geometric proof of this fact ?Thanks. – Erik Satie Apr 20 at 9:49

Your inequality is equivalent to : $$\sum_{cyc}\frac{a}{13}\sin(\arctan(\sqrt{\frac{13}{5}}\frac{a}{b}))^2\geq \frac{a+b+c}{18}$$ Each side is divided by $$b$$, We get: $$\frac{a}{13b} \sin(\arctan(\sqrt{\frac{13}{5}}\frac{a}{b}))^2+\frac{1}{13}\sin(\arctan(\sqrt{\frac{13}{5}}\frac{b}{c}))^2+\frac{c}{13b}\sin(\arctan(\sqrt{\frac{13}{5}}\frac{c}{a}))^2\geq \frac{1+\frac{a}{b}+\frac{c}{b}}{18}$$ Now we put $$\sqrt{\frac{13}{5}}\frac{a}{b}=x$$, $$\sqrt{\frac{13}{5}}\frac{b}{c}=y$$, $$\sqrt{\frac{13}{5}}\frac{c}{a}=z$$, your inequality is equivalent to: $$\sqrt{\dfrac{5}{13}}\frac{1}{13}\dfrac{(x)^3}{(x^2+1)}+\dfrac{1}{13}\dfrac{(y)^2}{(y^2+1)}+\sqrt{\dfrac{13}{5}}\dfrac{1}{13(y)}\dfrac{(z)^2}{(z^2+1)}$$$$\geq \dfrac{1+\sqrt{\dfrac{5}{13}}x+\sqrt{\dfrac{13}{5}}\dfrac{1}{y}}{18}$$ with the condition $$xyz=(\sqrt{\frac{13}{5}})^3$$.

We study the following function: $$f(x)=\sqrt{\dfrac{5}{13}}\frac{1}{13}\dfrac{(x)^3}{(x^2+1)}+\dfrac{1}{13}\dfrac{(y)^2}{(y^2+1)}+\sqrt{\dfrac{13}{5}}\dfrac{1}{13(y)}\dfrac{(z)^2}{(z^2+1)}-\dfrac{1+\sqrt{\dfrac{5}{13}}x+\sqrt{\dfrac{13}{5}}\dfrac{1}{y}}{18}$$ This function is easily differentiable and the minimum is for $$x=\sqrt{\frac{3\sqrt{77}}{17}-\frac{26}{17}}=\alpha$$. So with the condition $$xyz=(\sqrt{\frac{13}{5}})^3$$ becomes $$yz=\frac{(\sqrt{\frac{13}{5}})^3}{\alpha}=\beta$$. So we have this inequality just with $$y$$: $$\sqrt{\dfrac{5}{13}}\frac{1}{13}\dfrac{(\alpha)^3}{(\alpha^2+1)}+\dfrac{1}{13}\dfrac{(y)^2}{(y^2+1)}+\sqrt{\dfrac{13}{5}}\dfrac{1}{13(y)}\dfrac{(\frac{\beta}{y})^2}{((\frac{\beta}{y})^2+1)}\geq\dfrac{1+\sqrt{\dfrac{5}{13}}\alpha+\sqrt{\dfrac{13}{5}}\dfrac{1}{y}}{18}$$ which is easily analyzable. Done!

• Why the last inequality is true? – Michael Rozenberg Oct 21 '17 at 14:43
• Wow you analyze rapidly (without irony).I check it with wolfram alpha but I will create a question for this I think . – max8128 Oct 21 '17 at 14:45
• Delete please your "solution". I think it's not good, which you are doing. – Michael Rozenberg Oct 21 '17 at 16:59
• I think it's nothing. – Michael Rozenberg Oct 25 '17 at 11:15
• Why does this have 2 upvotes? The original inequality can also be analyzed via computer, which the author obviously did already. – Maxim Gilula Mar 15 at 10:46

A bit algebra shows that the inequality is equivalent to $$5 \left(5 a^5 \left(13 b^2+5 c^2\right)+13 a^4 \left(5 b^3-13 b^2 c-5 b c^2-5 c^3\right)+a^3 \left(-65 b^4+144 b^2 c^2+65 c^4\right)+a^2 \left(25 b^5-65 b^4 c+144 b^3 c^2+144 b^2 c^3-169 b c^4+65 c^5\right)-13 a \left(13 b^4 c^2+5 b^2 c^4\right)+5 b^2 c^2 \left(13 b^3+13 b^2 c-13 b c^2+5 c^3\right)\right) \ge0$$ The left hand side is a polynomial, so this can be solved by Cylindrical Algebra Decomposition -- https://en.wikipedia.org/wiki/Cylindrical_algebraic_decomposition

The following code in Mathematica does the job.

ex1 = a^3/(13 a^2 + 5 b^2) + b^3/(13 b^2 + 5 c^2) + c^3/(5 a^2 + 13 c^2) >= 1/18 (a + b + c);
ex2 = ex1[] - ex1[] // Together // Numerator // Simplify;
ex3 = ForAll[{a, b, c}, And @@ {a >= 0, b >= 0, c >= 0}, ex2 >= 0];
CylindricalDecomposition[ex3, {}]


Note this may take a few minutes to run.

I am not a reputable source, but I think I can prove the following theorem: $$\sum_{cyc}\frac{a_{1}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}\geq\frac{\sum_{k=1}^{n}a_{k}}{k_{1}+k_{2}}$$

PROOF

We will need firstly the following

Lemma 1.

$$\sum_{k=1}^{n}a_{k}^{\alpha+1}\geq\sum_{cyc}a_{1}a_{2}^{\alpha}$$

Proof.

Applying the Rearrangement inequality on $$a_{1},a_{2},...,a_{n}$$ and $$a_{1}^{\alpha},a_{2}^{\alpha},...,a_{n}^{\alpha}$$, we have that $$\sum_{cyc}a_{1}a_{2}^{\alpha}$$ is maximized when $$a_{1},a_{2},...,a_{n}$$ and $$a_{1}^{\alpha},a_{2}^{\alpha},...,a_{n}^{\alpha}$$ are similarly sorted.

Therefore, we can affirm that

$$\sum_{k=1}^{n}a_{k}^{\alpha+1}\geq\sum_{cyc}a_{1}a_{2}^{\alpha}$$

Other hand, we need the following

Lemma 2.

$$\sum_{cyc}\frac{a_{1}^{\alpha+1}+\left(\frac{k_{2}}{k_{1}}\right)a_{1}a_{2}^{\alpha}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}=\frac{\sum_{k=1}^{n}a_{k}}{k_{1}}$$

Proof.

We establish that

$$\frac{a_{1}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}+\frac{n}{m}=\frac{a_{1}}{k_{1}}$$

Operating, we get that

$$\frac{a_{1}^{\alpha+1}m+n\left(k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}\right)}{m\left(k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}\right)}=\frac{a_{1}}{k_{1}}$$

$$k_{1}\left(a_{1}^{\alpha+1}m+n\left(k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}\right)\right)=a_{1}m\left(k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}\right)$$

$$k_{1}a_{1}^{\alpha+1}m+k_{1}n\left(k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}\right)=a_{1}mk_{1}a_{1}^{\alpha}+a_{1}mk_{2}a_{2}^{\alpha}$$

$$k_{1}n\left(k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}\right)=a_{1}mk_{2}a_{2}^{\alpha}$$

$$\frac{n}{m}=\left(\frac{k_{2}}{k_{1}}\right)\frac{a_{1}a_{2}^{\alpha}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}$$

Therefore, we have that

$$\frac{a_{1}^{\alpha+1}+\left(\frac{k_{2}}{k_{1}}\right)a_{1}a_{2}^{\alpha}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}=\frac{a_{1}}{k_{1}}$$

And subsequently, repeating the process for each variable, we get that

$$\sum_{cyc}\frac{a_{1}^{\alpha+1}+\left(\frac{k_{2}}{k_{1}}\right)a_{1}a_{2}^{\alpha}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}=\frac{\sum_{k=1}^{n}a_{k}}{k_{1}}$$

Now, we are ready to prove the theorem.

Applying Lemma 1, we derive that

$$\left(\frac{k_{2}}{k_{1}}\right)\sum_{k=1}^{n}a_{k}^{\alpha+1}\geq\left(\frac{k_{2}}{k_{1}}\right)\sum_{cyc}a_{1}a_{2}^{\alpha}$$

Therefore, substituting in the expression of Lemma 2 and operating, we have that

$$\sum_{cyc}\frac{a_{1}^{\alpha+1}+\left(\frac{k_{2}}{k_{1}}\right)a_{k}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}\geq\frac{\sum_{k=1}^{n}a_{k}}{k_{1}}$$

$$\sum_{cyc}\frac{\left(\frac{k_{1}}{k_{1}}\right)a_{1}^{\alpha+1}+\left(\frac{k_{2}}{k_{1}}\right)a_{k}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}\geq\frac{\sum_{k=1}^{n}a_{k}}{k_{1}}$$

$$\sum_{cyc}\frac{\left(\frac{k_{1}+k_{2}}{k_{1}}\right)a_{k}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}\geq\frac{\sum_{k=1}^{n}a_{k}}{k_{1}}$$

$$\sum_{cyc}\frac{a_{k}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}\geq\frac{\sum_{k=1}^{n}a_{k}}{\left(\frac{k_{1}+k_{2}}{k_{1}}\right)k_{1}}$$

$$\sum_{cyc}\frac{a_{k}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}\geq\frac{\sum_{k=1}^{n}a_{k}}{k_{1}+k_{2}}$$

As we wanted to prove.

The particular case proposed follows from the direct application of the theorem.

• The first lemma is wrong because the right hand side is cyclic so you can't order the variables without losing generality. – LHF Mar 14 at 17:44
• You are right @Atticus! Do you think there could be some way to prove that $\sum_{cyc}a_{1}a_{2}^{\alpha}$ is maximum when all the variables are well ordered? That would fix the Lemma. – Juan Moreno Mar 14 at 18:11
• Maybe the Lemma could be proved using the Rearrangement inequality? In any case, for three variables, using the Rearrangement inequality it would be possible to prove that $a^3+b^3+c^3\geq ab^2+bc^2+ca^2$, and then the theorem would hold for cyclic sums of this type with three variables – Juan Moreno Mar 14 at 19:05
• The proof of Lemma 1 is ok. The generalized inequality you try to prove however is not true. For example $$\frac{a^8}{a^7+b^7}+\frac{b^8}{b^7+c^7}+\frac{c^8}{c^7+a^7}\geq \frac{a+b+c}{2}$$ does not hold for $(a,b,c)=(\frac{1}{4},\frac{1}{3},\frac{2}{5})$. Because of how you chose to write your proof, it's hard (for me) to pinpoint where exactly you went wrong. Sorry. – LHF Mar 14 at 21:46
• @LHF, I think I found the error... the theorem is only valid for variables greater than one, as $$\sum_{cyc}\frac{a_{1}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}\geq\sum_{cyc}\frac{\left(\frac{k_{2}}{k_{1}}\right)a_{1}a_{2}^{\alpha}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}$$ is not necesarily true for variables less than one. For variables greater than one, the theorem holds. – Juan Moreno Mar 17 at 16:07

Alex Ravsky prove the following statement :

Let $$a,b,c>0$$ such that $$a+b+c=1$$ and $$a\geq b\geq c$$ and $$13a^2+5b^2\geq 13b^2+5c^2\geq 13c^2+5a^2$$ then $$\exists n>1$$ such that :$$\frac{a^3}{n}+ \frac{(n-1)(13a^2+5b^2)}{54n}\geq \frac{13a^2+5b^2}{54}$$ $$\Big(\frac{a^3}{n}+ \frac{(n-1)(13a^2+5b^2)}{54n}\Big)\Big(\frac{b^3}{n}+ \frac{(n-1)(13b^2+5c^2)}{54n}\Big)\geq \frac{(13a^2+5b^2)(13b^2+5c^2)}{54^2}$$ $$\Big(\frac{a^3}{n}+ \frac{(n-1)(13a^2+5b^2)}{54n}\Big)\Big(\frac{b^3}{n}+ \frac{(n-1)(13b^2+5c^2)}{54n}\Big)\Big(\frac{c^3}{n}+ \frac{(n-1)(13c^2+5a^2)}{54n}\Big)\geq \frac{(13a^2+5b^2)(13b^2+5c^2)(13c^2+5a^2)}{54^3}$$

Or :

Let $$a,b,c>0$$ such that $$\frac{a^3}{13a^2+5b^2}\geq \frac{b^3}{13b^2+5c^2}\geq \frac{c^3}{13c^2+5a^2}$$ and $$a\geq b\geq c$$ and $$13a^2+5b^2\geq 13b^2+5c^2\geq 13c^2+5a^2$$ then $$\exists n>1$$ such that :$$\frac{\frac{a^3}{13a^2+5b^2}}{n}+ \frac{(n-1)(a+b+c)}{54n}\geq \frac{a+b+c}{54}$$ $$\Big(\frac{\frac{a^3}{13a^2+5b^2}}{n}+ \frac{(n-1)(a+b+c)}{54n}\Big)\Big(\frac{\frac{b^3}{13b^2+5c^2}}{n}+ \frac{(n-1)(a+b+c)}{54n}\Big)\geq \frac{(a+b+c)^2}{54^2}$$ $$\Big(\frac{\frac{a^3}{13a^2+5b^2}}{n}+ \frac{(n-1)(a+b+c)}{54n}\Big)\Big(\frac{\frac{b^3}{13b^2+5c^2}}{n}+ \frac{(n-1)(a+b+c)}{54n}\Big)\Big(\frac{\frac{c^3}{13c^2+5a^2}}{n}+ \frac{(n-1)(a+b+c)}{54n}\Big)\geq \frac{(a+b+c)^3}{54^3}$$

Remains to apply Karamata's inequality to get the desired result .By Karamata's inequality I mean this special case :

If $$a_1\geq a_2\geq a_3\geq\cdots\geq a_n$$ and $$b_1\geq b_2\geq b_3\geq\cdots\geq b_n$$ are two sequences of positive real numbers then we have $$\frac{a_1}{n}+\frac{(n-1)b_1}{n}\geq \frac{a_2}{n}+\frac{(n-1)b_2}{n} \geq\cdots\geq \frac{a_n}{n}+\frac{(n-1)b_n}{n}$$and $$b_1\geq b_2\geq b_3\geq\cdots\geq b_n$$ satisfying the following conditions(call the conditions $$C$$):$$\frac{a_1}{n}+\frac{(n-1)b_1}{n}\geq b_1,(\frac{a_1}{n}+\frac{(n-1)b_1}{n})(\frac{a_2}{n}+\frac{(n-1)b_2}{n})\geq b_1b_2,\cdots,(\frac{a_1}{n}+\frac{(n-1)b_1}{n})(\frac{a_2}{n}+\frac{(n-1)b_2}{n})\cdots (\frac{a_n}{n}+\frac{(n-1)b_n}{n})\geq b_1b_2\cdots b_n,$$ Then we have : $$a_1+a_2+a_3+\cdots+a_n\geq b_1+b_2+b_3+\cdots+b_n$$

Ps:If you look properly to the solution of Alex Ravsky we can invert $$a$$ and $$b$$ and get another case.

• No, because your inequality is obviously wrong. – Michael Rozenberg Mar 2 at 18:30
• @MichaelRozenberg I have checked my inequality is the same as yours and if you say that is wrong we got a contradiction in Math. – Erik Satie Mar 2 at 19:02
• @RiverLi what do you think about all of this ? – Erik Satie Mar 2 at 19:03
• @The.old.boy Since I did not comment here before, even you @ me, I did not know. I just looked at this post and see your comment. I do not know if your approach is correct. As I pointed out before, you only prove one of several cases. If you write the full solution for all cases, I may review it. – River Li Mar 13 at 7:13
• @The.old.boy If so, you should add explanation about it. – River Li Mar 13 at 8:48

I have finally found a solution . In fact we start to study the 2 variables version of this inequality we have :

$$\frac{a^3}{13a^2+5b^2}+\frac{b^3}{13b^2+5a^2}\geq \frac{a+b}{18}$$

Proof:

We have with $$x=\frac{a}{b}$$ : $$\frac{x^3}{13x^2+5}+\frac{1}{13+5x^2}\geq \frac{1+x}{18}$$ Or $$5(x+1)(x-1)^2(5x^2-8x+5)\geq 0$$

So we have (if we permute the variables $$a,b,c$$ and addition the three inequalities ) :

$$\sum_{cyc}\frac{a^3}{13a^2+5b^2}+\sum_{cyc}\frac{a^3}{13a^2+5c^2}\geq \frac{a+b+c}{9}$$

If we have $$\sum_{cyc}\frac{a^3}{13a^2+5b^2}\geq\sum_{cyc}\frac{a^3}{13a^2+5c^2}$$

We have : $$\sum_{cyc}\frac{a^3}{13a^2+5b^2}\geq \frac{a+b+c}{18}$$ But also $$\frac{(a-\epsilon)^3}{13(a-\epsilon)^2+5b^2}+\frac{(b)^3}{13(b)^2+5(c+\epsilon)^2}+\frac{(c+\epsilon)^3}{13(c+\epsilon)^2+5(a-\epsilon)^2}\geq \frac{a+b+c}{18}$$ If we put $$a\geq c$$ and $$\epsilon=a-c$$

We finally obtain : $$\sum_{cyc}\frac{a^3}{13a^2+5c^2}\geq \frac{a+b+c}{18}$$

So all the cases are here so it's proved !

• If you'll say that it's proved it's not says that it's indeed proved. Stop please to post a wrong solutions. – Michael Rozenberg Dec 7 '18 at 15:38
• Now it's right and proved . – max8128 Dec 8 '18 at 11:07
• I can only see that you proved that, for a given triple $(a, b, c)$, $\sum_{cyc}\frac{a^3}{13a^2+5b^2}\geq \frac{a+b+c}{18}$ or $\sum_{cyc}\frac{a^3}{13a^2+5c^2}\geq \frac{a+b+c}{18}$ must hold. – The same applies to your other answer math.stackexchange.com/a/3041504/42969. – Martin R Dec 15 '18 at 15:49