find minimum value of $n$ that satisfies this inequality: $4(x_1^2+x_2^2+...+x_n^2)<2(x_1+x_2+...+x_n)<x_1^3+x_2^3+...+x_n^3$

Question

I saw this question, it looks very hard:

there are real positive numbers $\{x_1,x_2,...x_n\}$ , given the inequality: $$4(x_1^2+x_2^2+...+x_n^2)<2(x_1+x_2+...+x_n)<x_1^3+x_2^3+...+x_n^3$$ What is the minimum value of $n$, for this to be possible?

This is what I've tried so far:

$$\sum_{k=1}^nx_k^3>\sum_{k=1}^n4x_k^2\\\sum_{k=1}^nx_k^3-\sum_{k=1}^n4x_k^2>0\\\sum_{k=1}^nx_k^2(x_k-4)>0$$So it must be at least one number that is bigger than $4$.

So I go with trial and error .Example 1;$$x_1=10, n=3001, x_2=x_3=...=x_{3001}=0.1\\4(100+3000\cdot0.01)=520<2(10+3000\cdot0.1)=620<1000+3000\cdot0.001=1003$$ Example 2;$$x_1=6, n=1001, x_2=x_3=...=x_{1001}=0.1\\4(36+1000\cdot0.01)=184<2(6+1000\cdot0.1)=212<216+1000\cdot0.001=217$$ S0 $n$ is getting smaller, but I have no idea how to minimize $n$, or even how to approach this kind of question.

Thanks in advance for any help.

Answer 1

It's 28th Tournament of the Towns, autumn 2006, A-Level, problem 7.

The answer is $516.$

An example for $n=516$.

Let $x_1=40k$, $x_2=x_3=...=x_{516}=k$, where $k>0$.

Thus, we need $$4(40^2+515)k^2<2(40+515)k<(40^3+515)k^3$$ or $$0.13116897...=\sqrt{\frac{74}{4301}}<k<\frac{37}{282}=0.1312056...$$ and $k=0.13117$ is valid.

Thus, it's enough to prove that for all $n\leq515$ our system has no solutions.

Let $n\leq515$ and $x_i$ be positive numbers such that $$4\sum\limits_{i=1}^nx_i^2<2\sum_{i=1}^nx_i<\sum_{i=1}^nx_i^3.$$ Also, let $x_i=ka_i$ such that $k>0$ and $$4\sum\limits_{i=1}^na_i^2=2\sum_{i=1}^na_i.$$ Thus, $$2k^2\sum_{i=1}^na_i^2<k\sum_{i=1}^na_i,$$ which gives $k^2<k$ or $0<k<1$ and we obtain $$\sum_{i=1}^nx_i^3>2\sum_{i=1}^nx_i$$ or $$k^2\sum_{i=1}^na_i^3>2\sum_{i=1}^na_i$$ and since $0<k<1$, we obtain $$\sum_{i=1}^na_i^3>2\sum_{i=1}^na_i,$$ which is a contradiction because we'll prove now that $$\sum_{i=1}^na_i^3\leq2\sum_{i=1}^na_i$$ for all $a_i>0$ such that $$2\sum\limits_{i=1}^na_i^2=\sum_{i=1}^na_i.$$ Indeed, we need to prove that $$\sum_{i=1}^na_i^3\leq2\sum_{i=1}^na_i\cdot\left(\frac{2\sum\limits_{i=1}^na_i^2}{\sum\limits_{i=1}^na_i}\right)^2$$ or $$8\left(\sum_{i=1}^na_i^2\right)^2\geq\sum_{i=1}^na_i^3\sum_{i=1}^na_i.$$

Thus, by the Vasc's EV Method, Corollary 1.8, case 3(b) here

https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06_www.pdf

it's enough to prove the last inequality for an equality case of $n-1$ variables and

since the last inequality is homogeneous, it's enough to assume $a_2=a_3=...=a_n=1$ and $a_1=x\geq1$.

Id est, we need to prove that $$8(x^2+n-1)^2\geq(x^3+n-1)(x+n-1)$$ or $$7(n-1)^2+(16x^2-x^3-x)(n-1)+7x^4\geq0,$$ which is obvious for $16x^2-x^3-x\geq0$ or for $(x^3+x-16x^2)^2-196x^4\leq0$.

Hence, it remains to prove it for $x^3+x>16x^2$ and $(x^3+x-16x^2)^2-196x^4\geq0$,

which is $x^3+x\geq30x^2$ or $x\geq15+4\sqrt{14}.$

Now, let $f(y)=7y^2+(16x^2-x^3-x)y+7x^4$.

Thus, since $$\frac{x^3+x-16x^2}{14}\geq x^2\geq(15+4\sqrt{14})^2>514,$$ we see that $f$ decreases on $[1,514]$ and it's enough to prove that $$7\cdot514^2+(16x^2-x^3-x)514+7x^4\geq0$$ or $$8(x^2+514)^2\geq(x^3+514)(x+514),$$ which is true because $$8(x^2+514)^2-(x^3+514)(x+514)>7\left(x^2-\frac{257x}{7}-152\right)^2\geq0.$$

Done!

Answer 2

This is a partial answer. Suppose that $m$ is the smallest positive integers such that there exists $x_1,x_2,\ldots,x_m>0$ for which $$4\,\sum_{i=1}^m\,x_i^2<2\,\sum_{i=1}^m\,x_i<\sum_{i=1}^m\,x_i^3\,.$$ Without loss of generality, let $x_1\leq x_2\leq \ldots\leq x_m$. We already know from other answers that $m\leq 516$. Obviously, $x_1<\dfrac12$ and $x_m>4$.

We note that $$2x_m^2-x_m<\sum_{i=1}^{m-1}\,x_i\,\left(1-2\,x_i\right)\leq \frac{m-1}{8}\leq \frac{515}{8}\,.$$ That is, $$x_m<\frac{1+2\sqrt{129}}{4}<5.93<6\,.$$

Now, for $t>0$, we have $$t(2x_m^2-x_m)-(x_m^3-2x_m)<\sum_{i=1}^{m-1}\,\Big(t(x_i-2x_i^2)-(2x_i-x_i^3)\Big)\,.\tag{*}$$ Define the function $f_t:\mathbb{R}\to\mathbb{R}$ by $$f_t(y):=t(y-2y^2)-(2y-y^3)\,.$$ Let $\tau:=\dfrac{106}{23}$. We note that $f_\tau$ is strictly decreasing on the interval $[4,6]$ and the maximum value of $f_\tau(y)$ for $y\in[0,6]$ is $\dfrac{61600}{328509}$, and this happens when $y=\dfrac{10}{69}$. That is, using (*), we obtain $$-f_\tau(4)<(m-1)\,f_\tau\left(\frac{10}{69}\right)\,,\text{ or }m>\frac{42959}{110}>390\,.$$ This proves that $m\geq 391$.

If I could prove that $x_m>5$, then it would follow that $m\geq 494$, which is quite an improvement. Better yet, if I knew that $x_m>5.2$, then I would get $m\geq 508$. If I used $f_T$ with $T:=\dfrac{3}{8}+\dfrac{1519}{32\sqrt{129}}$ instead, then I would improve the three inequalities by a little:

• $x_m>4$ implies $m\geq 395$;
• $x_m>5$ implies $m\geq 496$;
• $x_m>5.2$ implies $m\geq 511$.

If I study $f_s$ with $s:=\dfrac{73}{19}$ instead, then I can improve the bound for $x_m>4$ by a lot. That is, $$x_m>4\text{ implies }m\geq 461\,.$$ This is the best bound so far without extra assumptions on $x_m$, and I have no idea how to improve it (except very slightly, like $x_m>5$ leads to $m\geq 464$, which is not a very impressive improvement using $f_s$). Thus, so far, I can only say that $m\geq 461$, missing the intended target by $55$, but that may be improved if I can get a better bound for $x_m$.