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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.

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    $\begingroup$ 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. $\endgroup$ – Vincent May 8 '16 at 18:17
  • $\begingroup$ See also math.stackexchange.com/questions/1775572/… $\endgroup$ – David Quinn May 8 '16 at 18:42
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    $\begingroup$ @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. $\sum\limits_{cyc}\frac{a^3}{ka^2+b^2}\geq\frac{a+b+c}{k+1}$. $\endgroup$ – Michael Rozenberg Apr 19 '17 at 17:28
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    $\begingroup$ @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.. $\endgroup$ – Hanno May 11 '17 at 18:24
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    $\begingroup$ 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. $\endgroup$ – Hanno May 11 '17 at 18:50
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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!

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  • $\begingroup$ Why the last inequality is true? $\endgroup$ – Michael Rozenberg Oct 21 '17 at 14:43
  • $\begingroup$ Wow you analyze rapidly (without irony).I check it with wolfram alpha but I will create a question for this I think . $\endgroup$ – max8128 Oct 21 '17 at 14:45
  • $\begingroup$ We can check the starting inequality by WA without your previous work. But maybe WA is wrong (by the way, it happens sometimes!). $\endgroup$ – Michael Rozenberg Oct 21 '17 at 15:02
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    $\begingroup$ Delete please your "solution". I think it's not good, which you are doing. $\endgroup$ – Michael Rozenberg Oct 21 '17 at 16:59
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    $\begingroup$ I think it's nothing. $\endgroup$ – Michael Rozenberg Oct 25 '17 at 11:15
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With computer, I came up with two solutions, both of which are ugly.

Solution 1 (the SOS (Sum of Squares) method):

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 (by calculus):

WLOG, assume that $c = \min(a, b, c)$. WLOG, let $c = 1, \ b = 1+s, \ a = 1+t; \ s,t\ge 0.$ It suffices to prove that $f(s,t)\ge 0$ where $$f(s,t) = 25 s^5 t^2-65 s^4 t^3+65 s^3 t^4+65 s^2 t^5+50 s^5 t-135 s^4 t^2+351 s^2 t^4+130 s t^5+90 s^5-244 s^4 t-256 s^3 t^2+508 s^2 t^3+442 s t^4+90 t^5+216 s^4-928 s^3 t+504 s^2 t^2+496 s t^3+216 t^4+108 s^3-516 s^2 t+552 s t^2+108 t^3+72 s^2-72 s t+72 t^2.$$

Find the stationary points $\frac{\partial f}{\partial s} = \frac{\partial f}{\partial t} = 0$. Use the resultant of two polynomials. Ugly, omitted.

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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 !

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    $\begingroup$ If you'll say that it's proved it's not says that it's indeed proved. Stop please to post a wrong solutions. $\endgroup$ – Michael Rozenberg Dec 7 '18 at 15:38
  • $\begingroup$ Now it's right and proved . $\endgroup$ – max8128 Dec 8 '18 at 11:07
  • $\begingroup$ 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. $\endgroup$ – Martin R Dec 15 '18 at 15:49

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