I was wondering if we can always use Karamata's inequality in Olympiad problem .For that I use a special case of Karamata's inequality :
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 which satisfy the following conditions:$$a_1\geq b_1,a_1a_2\geq b_1b_2,a_1a_2a_3\geq b_1b_2b_3,a_1a_2\cdots a_n\geq b_1b_2\cdots b_n,$$ Then we have : $$a_1+a_2+a_3+\cdots+a_n\geq b_1+b_2+b_3+\cdots+b_n$$
Like that this statement works rarely so the idea is to use $n>1$ a natural number such that :
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 : $$a_1+a_2+a_3+\cdots+a_n\geq b_1+b_2+b_3+\cdots+b_n$$
The advantage of this new version is: the values are closer and the majorization works better .
Now we have the following statement :
Let $x_i>0$ and $y_i>0$ $(i=1,\cdots,n)$ be real numbers then we have : $$\sum_{i=1}^{n}\ln(y_i)-\sum_{i=1}^{n}\ln(x_i)\geq \sum_{i=1}^{n}\frac{y_i-x_i}{y_i}$$
So the conditions $C$ becomes $C_1$ :
$$\frac{\frac{a_1}{n}+\frac{(n-1)b_1}{n}-b_1}{\frac{a_1}{n}+\frac{(n-1)b_1}{n}}\geq0,\cdots,\sum_{k=1}^{n}\frac{\frac{a_k}{n}+\frac{(n-1)b_k}{n}-b_k}{\frac{a_k}{n}+\frac{(n-1)b_k}{n}}\geq 0,$$
If we want to solve the polynomials of variable $n$ ,it becomes hard for $n\geq 4$ the case $n=3$ is a second degree inequality .
Useless to say we don't solve Nesbitt's inequality with this method but I have tried (and I don't find a counter-example where $n$ doesn't exist ) we can solve : Inequality $\sum\limits_{cyc}\frac{a^3}{13a^2+5b^2}\geq\frac{a+b+c}{18}$
By example the last line of the majorization in the special case of Michael Rozenberg $(a,b,c)=(0.785,1.25,1.861)$ gives $n\geq 5224$
Futhermore I think we can solve the twin :Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$
My question :
What do you think about this method ? Is it already knew ?
Can someone show a hard inequality with that ?
Thanks a lot for your time and patience .
Edit :
Sketch of proof :
We show for $a,b,c>0$ : $$\frac{a^3}{13a^2+5b^2}+\frac{b^3}{13b^2+5c^2}+\frac{c^3}{13c^2+5a^2}\geq \frac{a+b+c}{18}$$ If we apply the version with $n$ we get :
Let $a,b,c>0$ such that $\frac{a^3}{13a^2+5b^2}\geq \frac{b^3}{13b^2+5c^2}\geq \frac{c^3}{13c^2+5a^2}$ then we have to show :$$\frac{\frac{a^3}{13a^2+5b^2}}{n}+ \frac{(n-1)(a+b+c)}{54n}\geq \frac{a+b+c}{54}$$ $$\Big(\frac{\frac{a^3}{13a^2+5b^2}}{n}+ \frac{(n-1)(a+b+c)}{54n}\Big)\Big(\frac{\frac{b^3}{13b^2+5c^2}}{n}+ \frac{(n-1)(a+b+c)}{54n}\Big)\geq \frac{(a+b+c)^2}{54^2}$$ $$\Big(\frac{\frac{a^3}{13a^2+5b^2}}{n}+ \frac{(n-1)(a+b+c)}{54n}\Big)\Big(\frac{\frac{b^3}{13b^2+5c^2}}{n}+ \frac{(n-1)(a+b+c)}{54n}\Big)\Big(\frac{\frac{c^3}{13c^2+5a^2}}{n}+ \frac{(n-1)(a+b+c)}{54n}\Big)\geq \frac{(a+b+c)^3}{54^3}$$
Or if we multiply by each denominator .
Let $a,b,c>0$ such that $a+b+c=1$ $a^3\geq b^3\geq c^3$ and $13a^2+5b^2\geq 13b^2+5c^2\geq 13c^2+5a^2$ then we have to show :$$\frac{a^3}{n}+ \frac{(n-1)(13a^2+5b^2)}{54n}\geq \frac{13a^2+5b^2}{54}$$ $$\Big(\frac{a^3}{n}+ \frac{(n-1)(13a^2+5b^2)}{54n}\Big)\Big(\frac{b^3}{n}+ \frac{(n-1)(13b^2+5c^2)}{54n}\Big)\geq \frac{(13a^2+5b^2)(13b^2+5c^2)}{54^2}$$ $$\Big(\frac{a^3}{n}+ \frac{(n-1)(13a^2+5b^2)}{54n}\Big)\Big(\frac{b^3}{n}+ \frac{(n-1)(13b^2+5c^2)}{54n}\Big)\Big(\frac{c^3}{n}+ \frac{(n-1)(13c^2+5a^2)}{54n}\Big)\geq \frac{(13a^2+5b^2)(13b^2+5c^2)(13c^2+5a^2)}{54^3}$$
The first line :
We need to show :
$$54a^3\geq 13a^2+5b^2$$
Or : $$54a^3\geq 18a^2\geq 13a^2+5b^2$$
Or :$$a\geq \frac{1}{3}$$
Second line :
We have to show using $C_1$:
$$\frac{\frac{-n}{54}}{\frac{a^3}{13a^2+5b^2}+\frac{n-1}{54}}+\frac{\frac{-n}{54}}{\frac{b^3}{13b^2+5c^2}+\frac{n-1}{54}}\geq -2$$
We put : $u=\frac{a^3}{13a^2+5b^2}$ and $v=\frac{b^3}{13b^2+5c^2}$
We have to show :
$$\frac{(108 uv + u n - 2 u + v n - 2 v - n/27 + 1/27)}{((54 u + n - 1) (54 v + n - 1))}\geq 0$$
Wich is obvious under the assumptions :
$$u>\frac{1}{54}, \frac{1}{54}<v<u, n>1 - 54 v$$
Third line :
We have to show using $C_1$:
$$\frac{\frac{-n}{54}}{\frac{a^3}{13a^2+5b^2}+\frac{n-1}{54}}+\frac{\frac{-n}{54}}{\frac{b^3}{13b^2+5c^2}+\frac{n-1}{54}}+\frac{\frac{-n}{54}}{\frac{c^3}{13c^2+5a^2}+\frac{n-1}{54}}\geq -3$$
We put :
$p=\frac{a^3}{13a^2+5b^2}$ and $q=\frac{b^3}{13b^2+5c^2}$ and $r=\frac{c^3}{13c^2+5a^2}$
As Wolfram alpha says remains to show that the roots exists .Remains to calculate a limit when the denominator is zero.Furthermore the expression under the root is positive see Inequality with a global minimum