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.

Solution 1:

@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 $$A_1 = \left(\begin{array}{ccccccc} 453245 & -100035 & 111397 & 113607 & -146718 & 24687 & 6498\\ -100035 & 166231 & -19773 & 24453 & -82004 & -11286 & 31356\\ 111397 & -19773 & 444600 & 86526 & -144261 & 13585 & -6669\\ 113607 & 24453 & 86526 & 760500 & -344736 & -126711 & 149188\\ -146718 & -82004 & -144261 & -344736 & 297882 & -26676 & -53352\\ 24687 & -11286 & 13585 & -126711 & -26676 & 160056 & -111150\\ 6498 & 31356 & -6669 & 149188 & -53352 & -111150 & 160056 \end{array}\right), $$ $$A_2 = \left(\begin{array}{cccccc} 281580 & 33098 & -64467 & 40261 & 4275 & -24219\\ 33098 & 329004 & -124982 & 13572 & 20241 & -51129\\ -64467 & -124982 & 106704 & -51129 & -6840 & -30875\\ 40261 & 13572 & -51129 & 326610 & -8645 & -79794\\ 4275 & 20241 & -6840 & -8645 & 62244 & -46683\\ -24219 & -51129 & -30875 & -79794 & -46683 & 153216 \end{array}\right) ,$$ $$A_3 = \left(\begin{array}{ccccccc} 258609 & -100035 & 102011 & 75582 & -56069 & -57798 & 126945\\ -100035 & 55575 & -37791 & -29393 & 2223 & 26923 & -26676\\ 102011 & -37791 & 589342 & 224757 & -195624 & -155376 & 313272\\ 75582 & -29393 & 224757 & 693576 & -279851 & -320283 & 535977\\ -56069 & 2223 & -195624 & -279851 & 200070 & 91143 & -306945\\ -57798 & 26923 & -155376 & -320283 & 91143 & 293436 & -435708\\ 126945 & -26676 & 313272 & 535977 & -306945 & -435708 & 1016158 \end{array}\right) ,$$ $$A_4 = \left(\begin{array}{ccccccc} 144495 & -57057 & 3705 & 24206 & -22230 & 2457 & 26923\\ -57057 & 55575 & 4199 & -15561 & -22724 & 13338 & -1989\\ 3705 & 4199 & 200070 & 18031 & -46449 & -9063 & 17784\\ 24206 & -15561 & 18031 & 351468 & -140049 & -6435 & -13509\\ -22230 & -22724 & -46449 & -140049 & 189202 & -117990 & -46449\\ 2457 & 13338 & -9063 & -6435 & -117990 & 177840 & -48659\\ 26923 & -1989 & 17784 & -13509 & -46449 & -48659 & 253422 \end{array}\right). $$

Solution 2:

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!

Solution 3:

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[[1]] - ex1[[2]] // 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.