Let $n$ be a given positive integer. Solve the system \[x_1 + x_2^2 + x_3^3 + \cdots + x_n^n = n,\] \[x_1 + 2x_2 + 3x_3 + \cdots + nx_n = \frac{n(n+1)}{2}\] in the set of nonnegative real numbers.
Problem
Source: Czech-Polish-Slovak 2005 Q1
Tags: algebra, polynomial, calculus, derivative, function, analytic geometry, graphing lines
djmathman
27.04.2013 21:28
Ahh, so this is where that question comes from.
LEMMA: For all integers $i$, the polynomial $f(x)=x^i-ix+i-1=0$ is nonnegative for all $x>0$.
Proof. First note that $f(1)=1-i+i-1=0$, so $x=1$ is a root. Taking the derivative of this function gives $f^\prime(x)=ix^{i-1}-i=i(x^{i-1}-1)$. When $0<x<1$, $x^{i-1}<1$, so the overall slope of the function is negative. At $x=1$, the slope of the function is $i(1-1)=0$, and when $x>1$, the slope of the function is positive. Therefore, the function can never dip below $x=1$ since it is an absolute minimum along the positive real numbers. $\blacksquare$
Subtracting the second equation from the first gives
\[x_2^2-2x_2+x_3^3-3x_3+\cdots+x_n^n-nx_n=n-\dfrac{n(n+1)}{2}=-\dfrac{n(n-1)}{2},\]
so
\begin{align*}x_2^2-2x_2+x_3^3-3x_3+2+\cdots+x_n^n-nx_n+(n-1)&=-\dfrac{n(n-1)}{2}+\left[1+2+\cdots+(n-1)\right]\\&=0.\end{align*}
This can be written more concisely as $\displaystyle\sum_{i=2}^n(x_i^i-ix_i+i-1)=0.$ From the Lemma, it is known that all the individual "terms" of this summation are nonnegative. When nonnegative expressions add up to $0$, each of them must be equal to $0$. Since it was established in the Lemma that $x=1$ is a root and a relative minimum of the expression $x_i^i-ix_i+i-1$ for all $i$, we have that $x_k=1$ for $k=2,3,\cdots,n$. Finally, plugging our known values into the first equation gives $x_1+\underbrace{1+1+\cdots+1}_{n-1\text{ terms}}=n$, so $x_1=1$ as well. Hence $\boxed{(x_1,x_2,\ldots,x_n)=(1,1,\ldots,1)}$ is the only solution to the system. $\blacksquare$
MexicOMM
05.10.2014 17:53
Sorry to revive but this problem looks exactly the same to the 1ยบ one of the Polish Finals of 2001. Strange knowing that Poland is in this contest. http://www.artofproblemsolving.com/Forum/viewtopic.php?p=376549&sid=fc2fe89563d10e6229e8d469a455ffb9#p376549
AHZOLFAGHARI
14.10.2014 22:26
$x_2^2 + 1 \geq 2x_2$ $x_3^3 + 1 +1 \geq 3 x_3$ . . . $x_n^n + 1 +1 + .. +1 \geq nx_n$ $x_1+x_2^2 + ... + x_n^n + \frac{n(n-1}{2}\geq nx^n + ... + x_1$ $ n + \frac{n(n-1}{2} \geq \frac{n(n+1}{2}$ but they are equal . so $ x_1 = x_2 = x_3 = ... = x_n = 1 $
cursed_tangent1434
10.01.2024 16:58
Note that, for all $1\leq i\leq n $, \[x_i^i +(i-1) \geq ix_i\]by AM-GM. Now summing gives us that, \[\sum_{i=1}^n x_i^i +\sum_{i=1}^n (i-1) \geq \sum_{i=1}^n ix_i\]Now, plugging in the values we are given yields, \[n+\frac{(n-1)n}{2} \geq \frac{n(n+1)}{2}\]So equality holds! But we know that equality holds in AM-GM if and only if all variables are equal from which we conclude that $x_1=x_2=\dots = x_n=1$ is the only solution as desired.
n2010
24.10.2024 12:54
Consider the following summations. By A.M-G.M Inequality, \[ x_1 = x_1\] \[ (x_2)^2 + 1\geq2x_2\] \[ (x_3)^3 + 1 + 1\geq3x_3 \] \[ \vdots \] \[ (x_n)^n + 1 + 1 + \ldots + 1 \text{(n-1 times)} \geq nx_n \] Adding all the above equations we get, \[ (x_1) + (x_2)^2 + \ldots + (x_n)^n + \frac{n}{2}(n-1) \geq (x_1 + 2x_2 + \ldots + nx_n) \] Putting the values given in the question, we get \[ \frac{n}{2}(n+1) \geq \frac{n}{2}(n+1) \] As it is the equality case, we have \( x_1 = x_2 = \ldots = x_n = 1 \) So the only solution to the system is \[ \boxed{(1, 1, \ldots, 1)} \].
AshAuktober
13.12.2024 15:54
Sol from my blog on Blogger: Okay. The first thing that we see is that $x_i = 1 \forall i$ works. Also I'd expect this to be the only solution; the equation feels very "restricted" in some sense. (I'm not sure how to formalise this intuition, so if anyone could help with that, that would be nice.) Also, the equations just seem to look like AM-GM? I'm not sure how exactly, but maybe I've done enough AM-GM ineq such as 2012 IMO 2 so that $x_i^i$ and $ix_i$ look like AM-GM. (Is this how geometers see configs?) Anyway, we'd be highly likely to find an AM-GM equality case from here. And equality holds for $x_i = 1$. So we're motivated (using Sharpness Principle, for example) to use AM-GM in the following way: $$x_i^i + (i-1) \ge i \left(x_i^i\right)^{\frac{1}{i}} $$ And summing this over all $i$, $n + \frac{n(n+1)}{2} - n \ge \frac{n(n+1)}{2}$. But equality holds, so equality holds everywhere, so $x_i = 1$ is indeed always true. $\square$ (One more thing to note is how nicely this ties in with nonnegative reals specifically)