Prove or Disprove this statement .

Question

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.

@MarkS. see here https://math.stackexchange.com/questions/3550018/a-special-case-of-karamatas-inequality-to-solve-one-or-more-olympiad-inequality . — Miss and Mister cassoulet char, Mar 01 '20 at 11:30
@Explorer It can be shown by a routine simplification of the expression and an application of a fact that $a\ge 1/3$ — Alex Ravsky, Mar 03 '20 at 21:29

Alex Ravsky · Accepted Answer · 2020-03-08T20:51:40.813

2

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 $m$.

edited Mar 08 '20 at 20:51

answered Mar 02 '20 at 16:45

Alex Ravsky

90,434

Could you tell me if the buffalo's way works as in my link ?Thanks for your answer. – Miss and Mister cassoulet char Mar 03 '20 at 10:53
Do you think we can use Muirhead's inequality ? – Miss and Mister cassoulet char Mar 03 '20 at 16:36
@The.old.boy For the inequality in the question? I think no, because Muirhead's inequality is for symmetric polynomials only. – Alex Ravsky Mar 03 '20 at 16:39
@The.old.boy I don’t remember what Buffalo way is, but formulas at the link are so complicated that I cannot see a way to a result from them. :-( – Alex Ravsky Mar 03 '20 at 21:28
1

I have tested your counterexample with the initial inequalities and it seems to work for $n=4000$ . Can you confirm ? Thanks .Furthermore i give you a link https://brilliant.org/discussions/thread/the-buffalo-way/ . – Miss and Mister cassoulet char Mar 04 '20 at 11:00
That's great indeed this factorization .I continue to work on this .Thanks a lot!! – Miss and Mister cassoulet char Mar 06 '20 at 13:53
Dear Alex Is it reasonable to publish these facts in a journal of Olympiad ? Maybe I'm dreaming =). – Miss and Mister cassoulet char Mar 07 '20 at 16:19
@The.old.boy First of all, the arguments are unfinished and the problem is unsolved. Next, especially taking into account the factorization, the problem is too hard for an olympiad, at least for a real-time that. When I was an olympiad winner I was good in inequalities, but this one is too hard and complicated for me even now and even with a computer assistance. Next, usually a solution of one inequality provides too few material to publish, even in arXiv. – Alex Ravsky Mar 08 '20 at 08:20
An opportunity is to look for a specific journal interested in such a stuff. Maybe “Crux Mathematicorum with Mathematical Mayhem” would fit, provided it is still published. Also this MSE thread can be the best publication place, because MSE is rather popular and this thread can be found by interested people via “Linked” and “Related” tabs or via Google search. – Alex Ravsky Mar 08 '20 at 08:20
Ah I was thinking the same thing for the Crux . But beyond this result we have a method to test others inequality of this kind like https://math.stackexchange.com/questions/1775572/olympiad-inequality-sum-limits-cyc-fracx48x35y3-geqslant-fracxy . I recognize that we have not finished and I have no ideas to achieve this . However I'm content to learn something with you . Thanks a lot . – Miss and Mister cassoulet char Mar 08 '20 at 10:14
1

I think that $r>0$ with the initial conditions of my problem so there is something wrong .I think we don't need this restriction .Are you agree ?And your last sentence is wrong . – Miss and Mister cassoulet char Mar 08 '20 at 16:26
Like this I cannot accept your answer because it miss some details of your calculations .Please add this . Thanks. – Miss and Mister cassoulet char Mar 09 '20 at 11:03
@The.old.boy As you wish. – Alex Ravsky Mar 09 '20 at 11:20
1

As I understand $r,u>0$ so we can conclude by the Archimedian property . But let me check your Buffalo's way . – Miss and Mister cassoulet char Mar 09 '20 at 13:58
1

The buffalo's way work !!! So I accept . Thanks a lot !!! – Miss and Mister cassoulet char Mar 09 '20 at 14:11
can i pick up some details of your proof for my answer here https://math.stackexchange.com/questions/1777075/inequality-sum-limits-cyc-fraca313a25b2-geq-fracabc18?noredirect=1&lq=1 ? – Miss and Mister cassoulet char Mar 17 '20 at 14:06
@The.old.boy Yes. – Alex Ravsky Mar 18 '20 at 03:01

Prove or Disprove this statement .

1 Answers1

Linked