# Polynomial inequality $n^2\sum_{i=1}^na_i^3\geq\left(\sum_{i=1}^na_i\right)^3$

Let $$n\ge 3$$ be an integer. I would like to know if the following property $$(P_n)$$ holds: for all real numbers $$a_i$$ such that $$\sum\limits_{i=1}^na_i\geq0$$ and $$\sum\limits_{1\leq i, we have $$n^2\sum_{i=1}^na_i^3\geq\left(\sum_{i=1}^na_i\right)^3.$$ I have a proof that $$(P_n)$$ holds for $$3\leq n\leq8$$, but for $$n\geq9$$ my method does not work and I did not see any counterexample for $$n\ge 9$$.

Is the inequality $$(P_n)$$ true for all $$n$$? Or otherwise, what is the largest value of $$n$$ for which it holds?

Thank you!

• @YCor I just think so because I solved during my live one problem or maybe two. Can I think so? You do not allow me? :) – Michael Rozenberg Apr 18 '20 at 14:16
• Well, it's confusing, as it conveys some wrong information. I edited your post; of course you can write that your guess is that it fails for large $n$, but "it seems" suggested that you have a serious reason to believe so. – YCor Apr 18 '20 at 15:07
• With Rolle (applied $n-3$ times to $\prod (x-a_i)$) it gives something which is wrong for large $n$. But I think Rolle does not reduce an inequality to the equivalent one: the antiderivative of a polynomial with real roots only may fail to have real roots only. I suggested something less elegant and straightforward: is three variables are mutually distinct, we may vary them so that the difference LHS-RHS increases. It reduces the problem to the situation when $a_i$'s take only two different values. – Fedor Petrov Apr 18 '20 at 18:21
• SageMath (actually just Python) gives $(-2, 1, 1, 1, 1, 1, 1, 1, 1)$ as a counterexample for $n=9$. Can you check? – darij grinberg Apr 18 '20 at 19:11
• @jcdornano yes: $6\sum a_i^3=(\sum a_i)^3-3(\sum a_i)(\sum a_i a_j)+3\sum a_i a_j a_k$. – Fedor Petrov Apr 18 '20 at 22:20

Take $$n=3k$$, $$2k$$ variables equal to $$3$$ and $$k$$ variables equal to $$-5$$ for large $$k$$. Then $$\sum a_i=k>0$$, and $$\sum_{i0$$ for large $$k$$. But $$\sum a_i^3<0$$.

• Thank you, Fedor! – Michael Rozenberg Apr 18 '20 at 20:01

This is just a long comment, but translating to the notation of symmetric functions, you ask if whenever $$e_{111}(x) \geq 0$$ and $$e_3(x) \geq 0$$, we have $$n^2 p_{(3)}(x) \geq p_{111}(x).$$ This latter is equivalent with $$n^2 \left( 3e_3-3e_{21}+e_{111} \right) \geq e_{111}.$$ Perhaps one can try different bases and see if something nice pops out...

A sort of (partial) explanation for what happens:

Let $$N \ge 3$$ the degree and let $$A=\sum{a_k}, B=\sum_{j. We are given that $$A \ge 0, C \ge 0$$ and we need to prove that $$N^2(A^3-3AB+3C) \ge A^3$$. Now we can assume wlog $$A =1$$ since if $$A=0$$ the inequality is obvious and otherwise we can divide by $$A>0$$ and consider $$c_j=\frac{a_j}{A}$$ and prove the inequality for them etc

So we need to prove $$1-3B+3C \ge \frac{1}{N^2}$$ under the hypothesis as above (the polynomial $$X^N-X^{N-1}+BX^{N-2}-CX^{N-3}+...$$ has real roots and $$C \ge 0$$

Then if we let $$b_j=a_j-\frac{1}{N}, A_1,B_1,C_1$$ the corresponding symmetric polynomials in $$b_j$$ we have $$A_1=0, B_1=B-\frac{N-1}{2N}=B-\frac {1}{2}+\frac{1}{2N}, C_1=C-B+\frac{2B}{N}+\frac{1}{3}-\frac{1}{N}+\frac{2}{3N^2}$$ so the inequality becomes ($$C-B=C_1+...$$ from the last equality)

$$1+3C_1-\frac{6B}{N}-1+\frac{3}{N}-\frac{2}{N^2}\ge \frac{1}{N^2}$$ and since

$$\frac{6B}{N}=\frac{6B_1}{N}+\frac{3}{N}-\frac{3}{N^2}$$ all reduces to

$$3C_1-\frac{6B_1}{N} \ge 0$$

But now the polynomial $$X^N+B_1X^{N-2}-C_1X^{N-3}+...$$ has real roots too as they are just $$b_k$$ and hence $$B_1 \le 0, B_1=-B_2, B_2 \ge 0$$ so the inequality reduces to $$2B_2+NC_1 \ge 0$$ and we know that $$C_1=C+\frac{N-2}{N}B_2-\frac{(N-1)(N-2)}{6N^2}$$

So we need $$C_1$$ negative but $$C \ge 0$$

By differentiating $$N-3$$ times and using Gauss Lucas/Rolle (so the cubic that results which is in standard form) has real roots so $$4(-p)^3 \ge 27q^2$$, we get some constraints on $$B_2, -C_1$$ which are enough to give the result for $$N \le 6$$ with some crude approximations

Then if we try easy counterexamples for the $$b_k$$ of the type $$N-1$$ $$a$$ and one $$-(N-1)a$$ we can solve at $$N=10$$, $$a > \frac{3}{80}$$ close enough to it to satisfy the inequality $$C>0$$ (which is satisifed at $$a=\frac{3}{80}$$ that one giving equality in the OP inequality as normalizing to integers we get $$11$$ taken $$9$$ times, $$-19$$ taken once and it is easy to see that the constraints are good and $$S_1=80, S_3=5120$$ and obviously $$100\cdot 5120=80^3$$

So as noted in the comments taking $$9$$ of $$111$$ and one of $$-199$$ gets a counterexample with a positive sum of cubes (corresponding to $$a=\frac{3}{80}+\frac{1}{800}$$ normalized to integers)