4
$\begingroup$

Let $a_i$ be distinct positive integers; prove that $$(a_1^7+a_2^7+\cdots + a_n^7)+(a_1^5+a_2^5+\cdots +a_n^5)\ge 2(a_1^3+a_2^3+\cdots + a_n^3)^2$$

I tried using some well known inequalities; obviously, since non homogenous and no obvious function, I don't expect either of AGM, Muirhead, CS, Jensen, Karamata, etc. should work, though I might be woefully wrong. After a while of experimentation I realized that this problem would likely be solved by either some tricky manipulations or a very obscure named inequality. Any helps? Thanks!

$\endgroup$
9
  • 1
    $\begingroup$ Are you able to do any special cases? Such as $n=1$, $n=2$? Anyway, it says distinct positive integers. Surely that is a key ingredient as the inequality does not hold for all positive real numbers. $\endgroup$ – Jyrki Lahtonen Sep 29 '20 at 4:20
  • 1
    $\begingroup$ Have you tried ordering $a_i$, and letting $d_i=a_{i+1}-a_i$? Feels like it might lead to somewhere. $\endgroup$ – David Cheng Sep 29 '20 at 4:24
  • 1
    $\begingroup$ @JJM Thanks, I am sorry for doubting you. I'd like the source anyway. $\endgroup$ – Teresa Lisbon Sep 29 '20 at 4:31
  • 2
    $\begingroup$ @TeresaLisbon np; its from a mock olympiad a friend gave me :) $\endgroup$ – user799557 Sep 29 '20 at 4:32
  • 1
    $\begingroup$ The case $n=1$ can be solved with AM-GM. Induction seems a good ideia to me. $\endgroup$ – Souza Sep 29 '20 at 4:54
1
$\begingroup$

One more induction proof

For $n=1$ everything is clear. Suppose that we know that inequality holds for $0<a_1<a_2<\dots<a_n$. We're going to add element $a_{n+1} > a_n$. Denote by $S_n^d = \sum_{i=0}^{n}a_i^d$ sum of $d$-th degrees of first $n$ elements. We know that $$S_n^7 + S_n^5 \ge 2(S_n^3)^2.\tag{1}$$ Adding $a_{n+1}$ gives $$ S_n^7 + S_n^5 + a_{n+1}^7 + a_{n+1}^5 \ge 2(S_n^3 + a_{n+1}^3)^2 $$ or, using $(1)$, $$ a_{n+1}^7 + a_{n+1}^5 \ge 4S_n^3a_{n+1}^3 + 2a_{n+1}^6 $$ $$ \frac{(a_{n+1} - 1)^2}{4a_{n+1}} \ge S_n^3a_{n+1}^{-3}\tag{2} $$ Now let's take a look at $S_n^3$. Since $a = a_{n+1} > a_n > \dots > a_0 > 0$, we have $$ S_n^3a_{n+1}^{-3} = \sum_{k=0}^{n}\left(\frac{a_k}{a_{n+1}}\right)^3 < \sum_{k=1}^{n+1}\left(1 - \frac{k}{a}\right)^3 < \sum_{k=1}^{a}\left(1 - \frac{k}{a}\right)^3 = \frac{(a - 1)^2}{4a} $$ which is exact left hand side of $(2)$.

$\endgroup$
4
  • $\begingroup$ Why is this true ? $\sum_{i=1}^{a_{n+1}}\left(1 - \frac{i}{a_{n+1}}\right)^3 = \frac{(a_{n+1} + 1)^2}{4a_{n+1}} - 1$ $\endgroup$ – user799557 Sep 29 '20 at 16:50
  • $\begingroup$ @JJM it's easy to see using induction $\endgroup$ – Anton Grudkin Sep 29 '20 at 19:27
  • $\begingroup$ Or you're asking something more precise?) $\endgroup$ – Anton Grudkin Sep 29 '20 at 19:28
  • 1
    $\begingroup$ My bad, i had misread oops. Thanks for your time. $\endgroup$ – user799557 Sep 29 '20 at 20:09
6
$\begingroup$

We can use induction here.

For $n=1$ it's true by AM-GM.

Now, let $$\sum_{k=1}^n(a_k^7+a_k^5)\geq\left(\sum_{k=1}^na_k^3\right)^2.$$ We'll prove that: $$\sum_{k=1}^{n+1}(a_k^7+a_k^5)\geq\left(\sum_{k=1}^{n+1}a_k^3\right)^2.$$ Indeed, let $a_{n+1}=a=\max\limits_k\{a_k\}$.

Thus, $$\sum_{k=1}^{n+1}(a_k^7+a_k^5)\geq \left(\sum_{k=1}^na_k^3\right)^2+a^7+a^5$$ and it's enough to prove that $$a^7+a^5\geq2a^6+4a^3\sum_{k=1}^na_k^3$$ or $$a^4+a^2\geq2a^3+4\sum_{k=1}^na_k^3.$$ Now, since $a_k\leq a-n-1+k,$ it's enough to prove that $$\sum_{k=1}^n(a-k)^3\leq\frac{1}{4}(a^4-2a^3+a^2)$$ or $$na^3-\frac{3n(n+1)}{2}a^2+\frac{3n(n+1)(2n+1)}{6}a-\frac{n^2(n+1)^2}{4}\leq\frac{1}{4}(a^4-2a^3+a^2)$$ or $$(a-n)^2(a-n-1)^2\geq0$$ and we are done!

As David Cheng said, we can prove the last inequality by the following simpler way. $$\sum_{k=1}^n(a-k)^3\leq\sum_{k=1}^{a-1}k^3=\frac{a^2(a-1)^2}{4}=\frac{1}{4}(a^4-2a^3+a^2).$$

$\endgroup$
5
  • $\begingroup$ You can use $\sum_{i=1}^{a-1} i^3=(\frac{a(a-1)}{2})^2=\frac 14 (a^4-2a^3+a^2)$ which make the proof simpler. $\endgroup$ – David Cheng Sep 29 '20 at 5:20
  • $\begingroup$ @David Cheng Did you mean $\sum\limits_{k=1}^n(a-k)^3\leq\sum\limits_{k=1}^{a-1}k^3=\frac{1}{4}(a^4-2a^3+a^2)$? $\endgroup$ – Michael Rozenberg Sep 29 '20 at 5:27
  • 1
    $\begingroup$ Yes, or alternatively $\sum_{k=1}^{n} a_k$ since they are less than $a$, and take distinct values. $\endgroup$ – David Cheng Sep 29 '20 at 5:30
  • $\begingroup$ @David Cheng It's nice! Thank you. I added it. $\endgroup$ – Michael Rozenberg Sep 29 '20 at 5:31
  • $\begingroup$ No problem at all. :) $\endgroup$ – David Cheng Sep 29 '20 at 5:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy