Let $n \in \mathbb{N}^*$ and $x_1,x_2,x_3,\ldots,x_n \in \mathbb{R}$' such that $0\le x_1\le x_2 \le \cdots \le x_n$, and $x_1+x_2+x_3+\cdots+x_n=1$, show that: $$(1+x_1^21^2)(1+x_2^22^2)\cdots(1+x_n^2n^2)\ge \frac{2n^2+9n+1}{6n}$$

I have no idea how you could solve this. I tried to manipulate the inequality using the inequality of arithmetic and geometric mean but it led me nowhere. I'm really interested in the solution of this problem. Any ideas?

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    $\begingroup$ Maybe I don't understand something but I don't think it's true for $x_1=x_2=\cdots=x_{n-1}=0$ and $x_n=1$. $\endgroup$ – pointer May 28 '14 at 16:07
  • $\begingroup$ Perhaps the inequality should be reversed? $\endgroup$ – Peter Woolfitt May 28 '14 at 16:15
  • $\begingroup$ Yes, sorry, now it's reversed. I double-checked the original problem text $\endgroup$ – Bardo May 28 '14 at 16:27
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    $\begingroup$ Emperically, the minimal value seems to be taken in the configuration $x_1=\cdots=x_n=\frac1n$. So maybe a two-step procedure? (1) Show that this configuration gives the minimal value (equivalently here, show that any other configuration has a nearby configuration that is smaller); (2) Establish the inequality for this configuration. Note also that the inequality constraints matter (the configuration $x_1=1$, $x_2=\cdots=x_n=0$ would give a counterexample), so order-blind tools like the AM/GM inequality won't immediately be relevant. $\endgroup$ – Greg Martin May 28 '14 at 17:52
  • $\begingroup$ I still don't know how to do it.. $\endgroup$ – Bardo May 28 '14 at 18:48

Consider two adjacent terms in the product, and use $x=x_j,y=x_{j+1}$ for the notation. The nondecreasing inequality then gives the constraint $x \le y$ and there is a positive integer $a=j^2$ so that the contribution to the product from the $j$ and $j+1$ terms is $$(1+a^2x^2)(1+(a+1)^2y^2). \tag{1}$$ Now if initially $x+y=k$ we only decrease (or keep the same) this two term product by replacing $x,y$ each by $k/2$, and note that since initially $x \le y$ this replacement has increased (or kept the same) $x$ while decreasing (or keeping the same) $y$, and so the replacement is consistent with the nondecreasing inequality among the $x_j$'s. It is also clear that, if $x\neq y$ were true initially, the replaced contribution would be strictly less than the initial product $(1).$

This argument shows that (as Greg Martin suggested) the minimal value of the expression occurs when all the $x_j$ are $1/n.$ If $p(n)=(2n^2+9n+1)/(6n)$ and $q(n)$ is the value of the product when each $x_j=1/n,$ then $p(1)=q(1)=2,$ so that for general $n$ the minimal lower bound of $p(n)$ happens to be exact at $n=1$. For larger $n$ it is less than $q(n)$, and has to be checked by hand for $n \le 4$, for which $q(2)-p(2)=1/4,$ and $q(3)-p(3)=53/81,$ and finally $q(4)-p(4)=655/512.$ [There may be a more clever way to see in general that $q(n)>p(n)$ but I don't see it.]

As soon as $n \ge 5$ we can use the fact that $1/n$ times the logarithm of the product is a right-endpoint sum for the integral $$\int_0^1 \log(1+x^2)\ dx = c= \log 2+(\pi-4)/2.$$ Then from $(1/n)\cdot \log q(n) \ge c$ we get $q(n) \ge e^{cn}.$ Though I didn't do this last part carefully, it seems that for $n \ge 5$ we do have $e^{cn} >p(n)$ as required. The left side here should certainly overtake the right since it's exponential with $e^c \approx 1.302.$ I guess the reason the integral bound isn't tight for $n<5$ is because too much error occurs in replacing the sum by the integral.

EDIT (at suggestion by the OP BarbuDorel in a comment). Once it is agreed the minimum of the product occurs when all the $x_j$ are $1/n$ the product becomes $$(1+(1/n)^2)(1+(2/n)^2)\cdots (1+(n/n)^2).$$ This is clearly at least $1$ plus the sum of the squared terms which occur as second terms in the factors. But $$1+(1/n)^2+(2/n)^2+\cdots+(n/n)^2=\frac{2n^2+9n+1}{6n},$$ on applying the formula for the sum of the first $n$ squares. So there is no need to consider integrals as I did above. [There still remains some work to justify replacing the product of two adjacent factors $(1)$ by the product obtained on replacing adjacent $x_j$ by their average. I'll add that if I can get a simple argument.

A simpler approach. Note that the given product is bounded below by the sum $$S=1+x_1^2+2^2x_2+\cdots + n^2x_n^2.$$ As noted above, if we can show this sum is minimal when all the $x_j$ are equal, the inequality follows. The argument for this goes similarly to before, looking at two adjacent terms, but the math is easier. Suppose the adjacent terms contribute $ax^2+by^2$ [where $0<a<b$] to the sum, and that initially $x<y$. Then if initially $x+y=k$ we have $x<k/2$. Now consider $$ax^2+b(k-x)^2=(a+b)x^2 -2bkx+bk^2.$$ The graph of this is a parabola opening upward, with its vertex at $x=bk/(a+b)>k/2,$ where the latter follows from $0<a<b.$ So since our initial $x$ satisfies $x<k/2$ we see that the value of $ax^2+by^2$ is strictly decreased when we replace each of $x,y$ by their average $(x+y)/2.$

Now since it is clear that the restrictions on the $x_k$ define a compact bounded region in $n$ space, the function $S$ must have a minimum, and the above argument shows this minimum cannot occur at any point at which any two of the $x_j$ are not equal. Conclusion: at the minimal $S$ all the $x_j$ are equal, and the overall inequality follows from that as just outlined.

  • $\begingroup$ It's a very clever and elegant argument. Honestly, I only understood about 70% of it, because I don't have the knowledge necessary to understand the part here you use the integral, but mostly even I got it. A very interesting thing yo point out would be that the problem book where I found this problem doesn't assume that the user has any knowledge of real-analysis, so I assume there is a simnpler proof. I'm still looking for it. What about induction? It was the first thing that crossed my mind, but I haven't been able to apply it properly. $\endgroup$ – Bardo May 30 '14 at 15:41
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    $\begingroup$ @BarbuDorel The part where the product of two terms called $(1)$ above is said to be lowered when $x+y=k$ and each is then replaced by $k/2$ was not proved in my answer. I have tried via basic calculus but haven't finished, will include if I do. The second part about using the integral is just to show the lower bound (when all $x_j$ equal) exceeds the formula on the right side. As I note, there may be a way without integrals to show this. Nice problem! [+1 on problem] $\endgroup$ – coffeemath May 30 '14 at 19:04
  • $\begingroup$ Ok, I found something quite interesting. The right-side term obviously reminded me of the sum of the sum $$ 1^2 + 2^2 + 3^2 + .. + n^2 = \frac{n(n+1)(2n+1)}{6} = \frac{2n^2+3n+1}{6} $$ . Now the right side-term is simply: $$\frac{2n^2+9n+1}{6n} = \frac{2n^2 + 3n + 1 + 6n}{6n} = \frac{2n^2+3n+1}{6n} + 1 = \frac{1}{n^2}\frac{2n^2+3n+1}{6} + 1 $$ And yes, I understood why you used the integral, thank you for the above explanation. Maybe this new-look on the right-side term will give us a new hint.. :-? $\endgroup$ – Bardo May 30 '14 at 19:13
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    $\begingroup$ @Bardo I found a different approach for which the relevant proof of minimality at all equal $x_j$ is much easier. It's based on considering only part of the expanded product. Let me know your reaction.. $\endgroup$ – coffeemath Jun 1 '14 at 2:34
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    $\begingroup$ Yes, I have just seen your argument, and I totally agree. It seems that our work here is done. Thank you very much for your help! $\endgroup$ – Bardo Jun 1 '14 at 15:04

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