Claim:
let $a,b,c>0$ and $p\geq 1$ then we have :
$$\left(\frac{a^3}{13a^2+5b^2}\right)^p+\left(\frac{b^3}{13b^2+5c^2}\right)^p+\left(\frac{c^3}{13c^2+5a^2}\right)^p\geq 3\left(\frac{a+b+c}{54}\right)^p$$
The case $p=1$ have been proved by user RiverLi and I offer a partial proof too in this case (see here).
The motivation:
In my partial proof (and a large part is due to user Alex Ravsky here) I use the well-know inequality called Karamata's inequality and Buffalo's way.
The fact is : the use of this inequality allow us to a generalisation because I apply the Karamata's inequality with the exponential function so it works for $e^x$ then it works also for $e^{xp}$.
Now and have a look to Alex's proof the only values for wich the inequality is valid is $p\geq 1$ .It's explained by another inequality called Jensen's inequality.
At the origin the inequality is due to Vasil Cirtoaje and strenghened by user Michael Rozenberg who found the coefficient $13$ and $5$.
Some details of the proof :
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 : $$a_1+a_2+a_3+\cdots+a_n\geq b_1+b_2+b_3+\cdots+b_n$$
To get the power $p$ we use Jensen's inequality because we have :
Let $u,v>0$ and $p$ a real number such that $p\geq 1$ and $n$ a natural number large enought:
$$u^p\frac{1}{n}+v^p\frac{n-1}{n}\geq \left(u\frac{1}{n}+v\frac{n-1}{n}\right)^p$$
To go further :
We have also a stronger statement :
let $a,b,c>0$ and $p\geq 1$ then we have :
$$\left(\frac{a^3}{13a^2+5b^2}\right)^p+\left(\frac{b^3}{13b^2+5c^2}\right)^p+\left(\frac{c^3}{13c^2+5a^2}\right)^p\geq \frac{a^p+b^p+c^p}{18^p}\geq 3\left(\frac{a+b+c}{54}\right)^p$$
To show it we come back at the end proof where I use Jensen's inequality .The inequality in $u,v$ can be strenghened using strong convexity and a modulus $m$ .I have not try but it seems promising .see here Theorem 2.
Using the idea above we got :Let $a,b,c>0 $ and $m=\min(\frac{a^{3}}{13a^{2}+5b^{2}},\frac{b^{3}}{13b^{2}+5c^{2}},\frac{c^{3}}{13c^{2}+5a^{2}},\frac{a+b+c}{54})$ and $p\geq 2$ a real number we have :
$$\left(\frac{a^{3}}{13a^{2}+5b^{2}}\right)^{p}+\left(\frac{b^{3}}{13b^{2}+5c^{2}}\right)^{p}+\left(\frac{c^{3}}{13c^{2}+5a^{2}}\right)^{p}-\left(3\left(\frac{\left(a+b+c\right)}{54}\right)^{p}+p\left(p-1\right)\left(m\right)^{\left(p-2\right)}\cdot0.5\cdot\left(\sum_{cyc}\left(\frac{\left(a+b+c\right)}{54}-\left(\frac{a^{3}}{13a^{2}+5b^{2}}\right)\right)^{2}\right)\right)\geq 0$$
Ps:The sign at the end is not a superior strict but superior or equal .
Question: How to show the claim properly?
