8
$\begingroup$

Show that for any non negative real numbers $x_1,x_2,\cdots x_n,$ $$\frac{{x_1}^2+{x_2}^2+\cdots+{x_n}^2}{n}x_1x_2\cdots x_n\le\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{n+2}$$

My work:

Let$$S(n)=\frac{{x_1}^2+{x_2}^2+\cdots+{x_n}^2}{n}x_1x_2\cdots x_n\le\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{n+2}$$ By theorem of triviality, if any of $x_i$'s are $0$ the inequality is certainly true. So assume all numbers are $\gt0$

$S(1)$ says ${x_1}^3\le {x_1}^3$ which is certainly true.

$S(2)$ says $({x_1}^2+{x_2}^2)x_1x_2\le\frac18(x_1+x_2)^4$ which reduces to $0\le(x_1-x_2)^4$ which is certainly true.

Assume $S(k)$ is true. Now we just needs to prove that $S(k+1)$ is true. But I'm having a hard time in doing that.

Any help is greatly appreciated. Or is there any better method than induction$?$

$\endgroup$
7
  • $\begingroup$ Have you tried maximizing the function after putting $k+1$ $\endgroup$
    – user1080568
    Aug 8, 2022 at 17:16
  • $\begingroup$ @GooglePlayGames can't understand ....pls elaborate $\endgroup$
    – abcdefu
    Aug 8, 2022 at 17:17
  • $\begingroup$ See this post. $\endgroup$ Aug 8, 2022 at 18:03
  • 1
    $\begingroup$ What is theorem of triviality? $\endgroup$ Aug 13, 2022 at 10:18
  • 1
    $\begingroup$ @TenaliRaman its just my way of writing "trivial solutions" $\endgroup$
    – abcdefu
    Aug 13, 2022 at 11:02

6 Answers 6

4
$\begingroup$

Let $k\ge2$ be fixed and suppose that $S(k)$ holds true. $$S(k):({x_1}^2+{x_2}^2+\cdots+{x_k}^2)x_1x_2\cdots x_k\le(k)\left(\frac{x_1+x_2+\cdots+x_k}{k}\right)^{k+2}$$ It remains to prove: $$S(k+1):({x_1}^2+{x_2}^2+\cdots+{x_k}^2+x^2)x_1x_2\cdots x_k\cdot x\le(k+1)\left(\frac{x_1+x_2+\cdots+x_k+x}{k+1}\right)^{k+3}$$ Using the fact that $x=x_{k+1}$

Now, Put $A=\frac{x_1+x_2+\cdots+x_k}{k}$ and $P=x_1x_2\cdots x_k$

$\implies$ $$S(k):({x_1}^2+{x_2}^2+\cdots+{x_k}^2)P\le kA^{k+2}$$ And it remains to prove that $$S(k+1):({x_1}^2+{x_2}^2+\cdots+{x_k}^2+x^2)Px\le (k+1)\left(\frac{kA+x}{k+1}\right)^{k+3}$$ The left hand side of $S(k+1)$ is $$({x_1}^2+{x_2}^2+\cdots+{x_k}^2)Px+Px^3\le kA^{k+2}x+Px^3$$ The above inequality came by using the fact of $S(k)$

So to prove $S(k+1)$, it suffices to prove that $$kA^{k+2}x+Px^3\le (k+1)\left(\frac{kA+x}{k+1}\right)^{k+3}$$ By $AM\ge GM$, $P\le A^k$, so it suffices to prove that $$kA^{k+2}x+A^kx^3\le (k+1)\left(\frac{kA+x}{k+1}\right)^{k+3}$$ Now restrict to the situation where the sum $x_1+x_2+\cdots+x_k+x$ is held constant, and prove the result with this added constraint. The general result then follows immediately; observe that for any constant $c$, the statement $S(n)$ holds for $x_1,\cdots,x_n$ if and only if it holds for $cx_1,\cdots,cx_n$ (the factor $c^{n+2}$ appears on each side). So, consider only those $(x_1,\cdots,x_k,x)\in\mathbb{R}^{k+1}$ for which $x_1+x_2+\cdots+x_k+x=k+1$,that is, $$kA+x=k+1$$ So, to prove $S(k + 1)$, it suffices to show $$kA^{k+2}x+A^kx^3\le k+1$$ The left hand of the above inequality is a function of $A$ (and $x = k + 1 − kA$, also a function of $A$), and so this expression is maximized using calculus: $$\frac{d}{dA}=[kA^{k+2}x+A^kx^3]=k(k+2)A^{k+1}x+kA^{k+2}\frac{dx}{dA}+kA^{k-1}x^3+A^k\cdot3x^2\frac{dx}{dA}$$ $$=k(k+2)A^{k+1}x+kA^{k-1}x^3-k(kA^{k+2}+A^k\cdot3x^2)$$ Putting $A = tx$, this expression becomes (after a bit of algebra) $$(1-t)(kt^2-2t+1)kt^{k-1}x^{k+2}$$ Since $k\ge2$, the above has roots at only $t = 0$ and $t = 1$, and so the derivative is positive for $0 < t < 1$ and negative for $t > 1$. Thus, $kA^{k+2}x+A^kx^3$ achieves a maximum when $t = 1$, that is, when $A = x = 1$. Hence, $$kA^{k+2}x+A^kx^3\le k+1$$ and so $S(k + 1)$ follows, completing the inductive step.

Thus, by mathematical induction, for all $n ≥ 1$, the statement $S(n)$ is true.

$\endgroup$
1
  • 2
    $\begingroup$ This is the work of many a person and not just mine; the credit goes to my fellas $\endgroup$
    – abcdefu
    Aug 13, 2022 at 16:12
4
+100
$\begingroup$

Simple AM-GM + Newton's solution is as follows: $$(x_1+x_2+\dots+x_n)^2 = \sum x_i^2 + \sum_{i\neq j}x_ix_j: = A + (n-1)B$$ with $A = \sum x_i^2\geq B = \dfrac{1}{n-1}\sum\limits_{i\neq j}x_ix_j,$ which is just pairwise AM-GM or rearrangement if you like. Then, $$\left(\dfrac{x_1+x_2+\dots +x_n}{n}\right)^{2n}=\left(\dfrac{A+(n-1)B}{n^2}\right)^n\geq \dfrac{1}{n^n}AB^{n-1}.$$ Using this, it suffices to prove then that: $$\left(\dfrac{B}{n}\right)^{n-1}\geq x_1x_2\dots x_n\left(\dfrac{x_1+x_2+\dots +x_n}{n}\right)^{n-2}.$$ In the language of elementary symmetric polynomials as you mentioned, this is written as: $$S_2^{n-1}\geq S_nS_1^{n-2}\iff \left(\dfrac{S_2}{S_1}\right)^{n-1}\geq\dfrac{S_n}{S_1} $$ But this follows from repeated applications of Newton's inequality: $$\dfrac{S_2}{S_1}\geq\dfrac{S_3}{S_2}\geq\dfrac{S_4}{S_3}\ \geq\dots \geq \dfrac{S_n}{S_{n-1}}\implies \left(\dfrac{S_2}{S_1}\right)^{n-1}\geq\prod_{k=1}^{n-1}\dfrac{S_{k+1}}{S_k} = \dfrac{S_n}{S_1}.$$

$\endgroup$
2
  • $\begingroup$ thanks for catching them. edited. $\endgroup$
    – dezdichado
    Aug 21, 2022 at 0:50
  • 1
    $\begingroup$ This is indeed a nice solution. +1. $\endgroup$
    – f10w
    Aug 21, 2022 at 13:34
4
$\begingroup$

WLOG, assume that $x_1+\dots+x_n = n$. We need to show that $f(x_1,\dots,x_n)\le n$, where \begin{equation} f(x_1,\dots,x_n) = x_1\dots x_n(x_1^2+\dots+x_n^2). \end{equation} Since the constraint set is closed and bounded, $f$ attains its maximum. In the following, we let $(x_1,\dots,x_n)$ denote such a maximum solution. Then, $x_i\neq 0 \forall i$, because otherwise $f = 0$, which is clearly not the maximum value.

Assume, to later arrive at a contradiction, that there exist two components of $(x_1,\dots,x_n)$ that are not equal to each other, say, $x_1\neq x_2$. Put $P = x_3\dots x_n, Q = x_3^2+\dots+x_n^2$, and $t=\frac{x_1+x_2}{2}$. Then we have \begin{align} f(x_1,\dots,x_n) - f(t,t,x_3,\dots,x_n) &= Px_1x_2(x_1^2+x_2^2+Q) - Pt^2(2t^2+Q)\\ &= -P\left[2t^4 - x_1x_2(x_1^2+x_2^2) + Q(t^2 - x_1x_2)\right]\\ &=-P\left[ \frac{1}{8}(x_1-x_2)^4 + \frac{Q}{4}(x_1-x_2)^2\right] < 0, \end{align} which is absurd as $(x_1,\dots,x_n)$ is a maximum solution. We conclude that $f$ achieves its maximum when $x_1=x_2=\dots = x_n$, QED.


Remarks.

  1. We have shown that $f(x_1,\dots,x_n) \le f(t,t,x_3,\dots,x_n)$, where $t=\frac{x_1+x_2}{2}$, which should immediately yield the conclusion using the classical mixing variable theorem, but the above proof by contradiction does not require this theorem.

  2. The same technique can be used to prove the following generalization: \begin{equation} (x_1\dots x_n)^p(x_1^q+\dots+x_n^q) \le n\left(\frac{x_1+\dots+x_n}{n}\right)^{np+q}, \end{equation} where $x_1,\dots,x_n\ge 0$ and $2p\ge \max\{q(q-1),0\}$.

$\endgroup$
4
  • $\begingroup$ It is nice. (Already upvoted 9 hours ago) $\endgroup$
    – River Li
    Aug 21, 2022 at 0:38
  • $\begingroup$ @RiverLi Thanks. I've just posted another solution by induction. Comments welcome! $\endgroup$
    – f10w
    Aug 21, 2022 at 13:33
  • $\begingroup$ Very nice (+1)! Is this approach generally applicable to all symmetric inequalities? Also nice generalization. Why must $2p\geq\max\{q(q-1),0\}$? $\endgroup$
    – V.S.e.H.
    Aug 31, 2022 at 8:28
  • 1
    $\begingroup$ @V.S.e.H. Thanks. This method is definitely not applicable to all symmetric inequalities, not even close. Sometimes one needs to resort to a stronger form: $f(x_1,\dots,x_n) \le \max\{f(t,t,x_3,\dots,x_n), f(0,2t,x_3,\dots,x_n)\}$. This can cover a much larger sets of inequalities (but of course still far from all symmetric inequalities). Regarding the generalization, the condition on $p$ and $q$ is such that the inequality holds for $n=2$. It's likely that this can be made stronger but I didn't want to spend much time on it. $\endgroup$
    – f10w
    Aug 31, 2022 at 10:30
1
$\begingroup$

Another simple solution by induction, motivated by OP's questions and by the currently most voted answer (which is also by induction but a bit complicated).

\begin{equation} \frac{{x_1}^2+{x_2}^2+\cdots+{x_n}^2}{n}x_1x_2\cdots x_n\le\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{n+2} \end{equation} First, the inequality is clearly true for $n=2$, as it can be reduced to $(x_1-x_2)^4\ge 0$. Denote $p = x_1\dots x_n, q = (x_1^2+\dots + x_n^2)/n, s = (x_1+\dots+x_n)/n$, and $x=x_{n+1}$. Assume that the inequality is true for $n$ numbers, that is, $s^{n+2} \ge pq$. We will show that it is also true for $n+1$ numbers, that is, $f(x) \ge 0$, where \begin{equation} f(x) = \left(\frac{x+ns}{n+1}\right)^{n+3} - \frac{px(x^2 + nq)}{n+1}. \end{equation} WLOG, assume that $x_{n+1}$ is the maximum component. Then, it suffices to show that $f(x) \ge 0$ whenever $x\ge s$. Taking derivatives: \begin{align} f'(x) &= \frac{n+3}{n+1}\left(\frac{x+ns}{n+1}\right)^{n+2} - \frac{p(3x^2 + nq)}{n+1},\\ f''(x) &= \frac{(n+3)(n+2)}{(n+1)^2}\left(\frac{x+ns}{n+1}\right)^{n+1} - \frac{6px}{n+1}. \end{align} We can observe that $f(s)\ge 0$ and $f'(s) \ge 0$. Indeed, as $s^{n+2} \ge pq$ and $s^n\ge p$ (AM-GM), we have \begin{align} f(s) &= s^{n+3} - \frac{ps(s^2 + nq)}{n+1} = \frac{ns(s^{n+2}-pq) + s^3(s^n-p)}{n+1} \ge 0,\\ f'(s) &= \frac{(n+3)s^{n+2}}{n+1} - \frac{p(3s^2 + nq)}{n+1} = \frac{n(s^{n+2}-pq)}{n+1} + \frac{3s^2(s^n-p)}{n+1} \ge 0. \end{align} Next, notice that $\left(\frac{x+ns}{n+1}\right)^{n+1} \ge xs^n \ge xp$ (AM-GM), we have \begin{equation} f''(x) \ge \frac{(n+3)(n+2)px}{(n+1)^2} - \frac{6px}{n+1} = \frac{n(n-1)px}{(n+1)^2} \ge 0. \end{equation} Therefore, $f'$ is increasing (on $[s,+\infty)$) and thus $f'(x) \ge f'(s) \ge 0$, hence $f$ is increasing and thus $f(x)\ge f(s) \ge 0$. QED.

$\endgroup$
0
$\begingroup$

theorem :

Let $x_1,x_2,x_3,y_1,y_2,y_3\in(0,\infty)$ then if we have :

$$x_1+x_2+x_3\geq y_1+y_2+y_3$$

And for $i\neq j,1\leq i\leq 3,1\leq j\leq 3$:

$$|x_i-x_j|\leq |y_i-y_j|$$

Then we have :

$$x_1x_2x_3\geq y_1y_2y_3$$

Case $n=3$ :

Let $y_1=\left(3-\frac{6\left(ab+bx+xa\right)}{\left(a+b+x\right)^{2}}\right)^{\frac{1}{3}}\sqrt{xa},y_2=\left(3-\frac{6\left(ab+bx+xa\right)}{\left(a+b+x\right)^{2}}\right)^{\frac{1}{3}}\sqrt{xb},y_3=\left(3-\frac{6\left(ab+bx+xa\right)}{\left(a+b+x\right)^{2}}\right)^{\frac{1}{3}}\sqrt{ab}$

Let : $x_1=x_2=x_3=\left(a+b+x\right)/3$

We have the easy inequalities :

$$0\geq\left(3-\frac{6\left(\sqrt{xa}+\sqrt{xb}+\sqrt{ab}\right)}{3\left(a+b+x\right)}\right)^{\frac{1}{2}}\left(\sqrt{xa}+\sqrt{xb}+\sqrt{ab}\right)-\left(a+b+x\right)\geq \left(3-\frac{6\left(ab+bx+xa\right)}{\left(a+b+x\right)^{2}}\right)^{\frac{1}{3}}\left(\sqrt{xa}+\sqrt{xb}+\sqrt{ab}\right)-\left(a+b+x\right)$$

Remains to apply the theorem to get the case $n=3$ with a refinement (bonus)

The general case is similar .Also think to Maclaurin's inequalities .

Some details :

A classical idea is using the Bernoulli's inequality as we have with $a,b,c>0$ :

$$-2\left(2-\frac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^{2}}\right)+3\left(2-\frac{2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}{\left(a+b+c\right)}\right)\leq \left(3-\frac{6\left(\sqrt{ca}+\sqrt{cb}+\sqrt{ab}\right)}{3\left(a+b+c\right)}\right)^{3}-\left(3-\frac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^{2}}\right)^{2}$$ $$

$\endgroup$
-1
$\begingroup$

My try :

Lemma :

Let $x>0$ then define :

$$f(x)=\ln(x)$$

Then a trivial consequence is :

$$f'''(x)>0$$

Now the problem is with the constraint of the OP:

$$\ln\left(\frac{{x_1}^2+{x_2}^2+\cdots+{x_n}^2}{n}\right) +\ln(x_1)+\ln(x_2)+\cdots +\ln(x_n)-\ln\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{n+2}\leq 0$$

Now using the lemma above and the Equal Variable Method with the constraint :

$$x_1+x_2\cdots+x_n=constant,x_1^2+x_2^2\cdots+x_n^2=constant$$

You can maximize the function and get another inequality really simpler

Edit after RiverLi's comment :

We need to show for $n\geq 3$ an integer $x,a>0$:

$$f(x)=\frac{\left(\left(n-1\right)\left(\frac{x}{a}\right)^{2}+1\right)}{n}\left(\frac{x}{a}\right)^{\left(n-1\right)}-\left(\frac{\left(n-1\right)x\cdot\frac{1}{a}+1}{n}\right)^{\left(n+2\right)}\leq 0\tag{E}$$

we have taking logarithm on both side in $E$ and differentiating where $a=1$:

$$h(x)=\frac{\left(n-1\right)\left(a-x\right)\left(a^{2}-2ax+\left(n-1\right)x^{2}\right)}{x\left(a+\left(n-1\right)x\right)\left(a^{2}+x^{2}\left(n-1\right)\right)}$$

For $0<x\leq a $ and $n\geq 2$ we have the following trivial inequalities :

$$a-x\geq 0$$

And :

$$a^{2}-2ax+\left(n-1\right)x^{2}\geq 0$$

So the derivative is positive and the function is increasing .Now remarking for $x=a$ we get zero as value the inequality is established .

$\endgroup$
17
  • $\begingroup$ Can you provide a detailed proof? Although Equal Variable Theorem works, the proof is not very easy. $\endgroup$
    – River Li
    Aug 11, 2022 at 12:20
  • 1
    $\begingroup$ I hope to see your complete detailed proof. It seems you never give a complete detailed proof. $\endgroup$
    – River Li
    Aug 11, 2022 at 15:05
  • 2
    $\begingroup$ By the way, I have two proofs. So I just want to see if you can give a complete detailed proof. $\endgroup$
    – River Li
    Aug 11, 2022 at 15:07
  • 3
    $\begingroup$ Waiting for your complete detailed proof. Otherwise it is not an answer, I think. $\endgroup$
    – River Li
    Aug 12, 2022 at 0:27
  • 1
    $\begingroup$ @SuzuHirose My proofs are not nice. I will not post it if there are other proofs, unless my proofs are better. abcdefu has given a nice proof (6 upvotes). Thanks for your attention. $\endgroup$
    – River Li
    Aug 18, 2022 at 9:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .