Partial Answer
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.
The easy case as $a\geq b \geq c>0$ :
let $x\in[0,1]$ then define :
$$h\left(x\right)=\frac{1}{9}\left(1+\left(\frac{3}{2+x}-1\right)\frac{\left(1+x\frac{4.5+\frac{4.75x}{2}\left(1-x\right)}{3}\right)}{3}\right)$$
Then we have $x\in[0,1]$:
$$f\left(x\right)=\frac{1}{13+5x^{2}}+\frac{1}{18}-h\left(x\right)\geq 0$$
Remains to show $a\geq b \geq c>0$:
$$\frac{c^{3}}{13c^{2}+5a^{2}}-\frac{2\left(a+b\right)+c}{18}+\left(ah\left(\frac{b}{a}\right)+bh\left(\frac{c}{b}\right)\right)\geq 0\tag{I}$$
Dividing by $a$ $(I)$ and plugging $X=\frac{b}{a}$ and $Y=\frac{c}{b}$ we have :
$$g(X,Y)=Xh(Y)+h(X)-(X+1)\cdot\frac{1}{9}-\frac{XY}{18}+XY\cdot\frac{1}{13+\frac{5}{\left(XY\right)^{2}}}$$
Using the fact that for $Y$ fixed we have :
Let $0<x\leq y\leq 1$ :
$$h(x)\geq h(y)$$
And for $0<x\leq y\leq 1$ and $\frac{1}{3}\leq a\leq 1$:
$$xh\left(a\right)-\left(\frac{x+1}{9}\right)-\frac{xa}{18}+\frac{xa}{18}\cdot\frac{1}{13+\frac{5}{\left(xa\right)^{2}}}\geq yh\left(a\right)-\left(\frac{y+1}{9}\right)-\frac{ya}{18}+\frac{ya}{18}\cdot\frac{1}{13+\frac{5}{\left(ya\right)^{2}}} $$
To show this two fact one can use the definition of a decreasing function .
The rest is smooth because now two variables on $a,b,c$ are equal .
For the hard case we have an interesting result with :
$$H(x)=\frac{2}{18}\left(1+\left(1+\frac{1.5x^{2}\left(x-1\right)^{2}}{\left(x+1\right)^{2}}\right)\left(\frac{3}{2+x}-1\right)\frac{\left(1+x\frac{4.5+\frac{4.75x}{2}\left(1-x\right)}{3}\right)}{3}\right)$$
To be continued ...
Third way :
It seems that the function for $u\geq k\geq p>0$ :
$$\frac{x}{13+5\left(\frac{p\left(x+u\right)}{x+k}\right)^{2}}=f(x)$$
is convex on $(0,\infty)$ so using the substitution :
$$b=a\left(a+u\right)\cdot\frac{p}{a+k},c=b\left(b+u\right)\cdot\frac{p}{b+k}$$
Remains to show :
$$2f\left(\frac{\left(a+b\right)}{2}\right)+\frac{c^{3}}{13c^{2}+5a^{2}}-\frac{a+b+c}{18}\geq 0$$
Wich seems true .
To be continued .
Edit 17/03/2022 :
Taking $d>0$ such that :
$$\frac{c^{3}}{13c^{2}+5a^{2}}=f\left(d\right)$$
And using the same strategy as above (where we use the convexity of $f(x)$ and Jensen's inequality :
We have to show :
$$3f\left(\frac{\left(a+b+d\right)}{3}\right)-\frac{\left(a+b+c\right)}{18}\geq 0$$
Remark : To lighten the cubic to a second degree polynomial we can use the substitution $d=ct$ and then use weighted Jensen's inequality .
Second remark :
We can introduce the function :
$$g\left(x\right)=\frac{c}{13+\frac{5a^{2}}{c^{2}}}-\frac{c}{13+\frac{5p^{2}\left(x+u\right)^{2}}{\left(x+k\right)^{2}}}$$
Now it's not a quadratic but we have to solve a one degree polynomial or $g(d)=0$. Where $d>0$
Unfortunetaly it adds some constraint .
In fact with the constraint above the inequality is proved for :
$$p\leq \frac{k+\frac{\left(a+b+c\right)}{1+1+\frac{c}{d}}}{u+\frac{\left(a+b+c\right)}{1+1+\frac{c}{d}}}$$
Last edit 18/03/2022:
In fact there is a contradiction with the last constraint with $p$ . It should be :
$$a=\left(\frac{bp\left(b+u\right)}{b+k}\right),b=c\left(\frac{c\left(c+u\right)}{c+k}\right)$$
And :
$$r\left(x\right)=\frac{a}{13+\frac{5c^{2}}{a^{2}}}-\frac{a}{13+\frac{5p^{2}\left(x+u\right)^{2}}{\left(x+k\right)^{2}}}$$
$$\frac{\frac{\left(a+b+c\right)}{2+\frac{a}{d}}+k}{\frac{\left(a+b+c\right)}{2+\frac{a}{d}}+u}\geq p$$
A computer algebra system reduces this problem to a system of polynomial equations and then solves this system by computing a Gröbner basis and numerically approximating the roots of polynomials with very high degrees.
What kind of proof are you looking for?
– cafaxo Apr 13 '17 at 14:23