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](https://math.stackexchange.com/questions/1777075/inequality-sum-limits-cyc-fraca313a25b2-geq-fracabc18?noredirect=1&lq=1)).




---


The motivation:
---------------


In my partial proof (and a large part is due to user Alex Ravsky [here](https://math.stackexchange.com/questions/3563335/prove-or-disprove-this-statement?noredirect=1&lq=1)) I use the well-know inequality called [Karamata's inequality](https://en.wikipedia.org/wiki/Karamata%27s_inequality) and [Buffalo's way](https://brilliant.org/discussions/thread/the-buffalo-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](https://journalofinequalitiesandapplications.springeropen.com/articles/10.1186/s13660-018-1897-2#:%7E:text=The%20Jensen%20inequality%20for%2) 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?



Olympiad inequality as a generalizing result due at the origin to Vasile Cirtaoje
The case $p>1$ easily follows from the case $p=1$. Indeed, let $M\_p(x,y,z) := \left(\frac{x^p+y^p+z^p}3\right)^{1/p}$. Then for $p>1$ we have
$$M\_p\big( \frac{a^3}{13a^2+5b^2}, \frac{b^3}{13b^2+5c^2}, \frac{c^3}{13c^2+5a^2}\big) \stackrel{(1)}{\geq} M\_1\big( \frac{a^3}{13a^2+5b^2}, \frac{b^3}{13b^2+5c^2}, \frac{c^3}{13c^2+5a^2}\big) \stackrel{(2)}{\geq} \frac{a+b+c}{54},$$
where (1) the [power mean inequality](https://en.wikipedia.org/wiki/Generalized_mean#Generalized_mean_inequality) and (2) is the case $p=1$.


