I can't find a counter-example to 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 :$$\frac{a^3}{n}+ \frac{(n-1)(13a^2+5b^2)}{54n}\geq \frac{13a^2+5b^2}{54}$$ $$\Big(\frac{a^3}{n}+ \frac{(n-1)(13a^2+5b^2)}{54n}\Big)\Big(\frac{b^3}{n}+ \frac{(n-1)(13b^2+5c^2)}{54n}\Big)\geq \frac{(13a^2+5b^2)(13b^2+5c^2)}{54^2}$$ $$\Big(\frac{a^3}{n}+ \frac{(n-1)(13a^2+5b^2)}{54n}\Big)\Big(\frac{b^3}{n}+ \frac{(n-1)(13b^2+5c^2)}{54n}\Big)\Big(\frac{c^3}{n}+ \frac{(n-1)(13c^2+5a^2)}{54n}\Big)\geq \frac{(13a^2+5b^2)(13b^2+5c^2)(13c^2+5a^2)}{54^3}$$
Pari-Gp have run and there is nothing against this statement .But I have a doubt .The first line is obvious .So my question concern only the two others .
If someone could prove or disprove this it will be cool.
Thanks a lot to your time .
Edit : If it's works we can add to my reasoning the Buffalo's way like here.
Put $N=54$, $m=n-1>0$, $x=13b^2+5c^2$, $y=13c^2+5a^2$, and $z=13a^2+5b^2$. Then the second inequality transforms to $rm+s\ge 0$, where $r=N(a^3x+b^3z)-2xz$ and $s=N^2a^3b^3-xz$. The third inequality transforms to $um^2+vm+w\ge 0$, where $u=N(a^3xy+b^3yz+c^3xz)-3xyz$, $v=N^2(a^3b^3y+a^3c^3x+b^3c^3z)-3xyz$, and $w=N^3a^3b^3c^3-xyz$.
Now it’s time to follow the Buffalo way. Put $b=c+p$, $a=c+p+q$, and replace $xz$ by $(a+b+c)xz$ to make the total degrees of all monomials equal. When we open the brackets in the expressions for $r$ and $u$ and simply (I did this with Mathcad), we obtain long sums of products of non-negative numbers with positive coefficients. Each of the sums equal to zero iff $p=q=0$. Thus if $p=q=0$ then $a=b=c$, $x=y=z=18a^2$, so $r=s=u=v=w=0$ and both inequalities become equalities. Otherwise $r,u>0$ and both inequalities are satisfied for sufficiently big $m$.