Given is a real number $a \in (0,1)$ and positive reals $x_0, x_1, \ldots, x_n$ such that $\sum x_i=n+a$ and $\sum \frac{1}{x_i}=n+\frac{1}{a}$. Find the minimal value of $\sum x_i^2$.
Problem
Source: All-Russian MO 2023 Final stage 10.8
Tags: inequalities, algebra
S.Das93
24.04.2023 06:25
By Cauchy-Schwarz we have $(\sum x_i^2)(\sum 1) \ge (\sum x_i)^2$. Simplifying gives $\sum x_i^2 \ge \frac{n^2+an+a^2}{n+1}$. Although I'm not sure if this is actually the minimum value...
SneedsFeedAndSeed
24.04.2023 11:07
a_507_bc wrote:
The equality case is $(1,1, \ldots, 1, a)$.
Do you have a solution?
EthanWYX2009
24.04.2023 12:50
a_507_bc wrote: Given is a real number $a \in (0,1)$ and positive reals $x_0, x_1, \ldots, x_n$ such that $\sum x_i=n+a$ and $\sum \frac{1}{x_i}=n+\frac{1}{a}$. Find the minimal value of $\sum x_i^2$. We have $$(x_0-a)+\sum\limits_{k=1}^n(x_k-1)=0,$$$$(\frac 1{x_0}-\frac 1a)+\sum\limits_{k=1}^n\left(\frac 1{x_k}-1\right)=0.$$Then $$\begin{aligned} \sum x_i^2-n-a^2 &=(x_0^2-a^2)+\sum\limits_{k=1}^n(x_i^2-1) \end{aligned}$$I'm not sure but maybe it can be done by Chebyshev.
RagvaloD
24.04.2023 14:09
For $x_0=a,x_i=1$ we get value $n+a^2$. Let show, that it is minimal
$\sum x_k^2=\sum (1-x_k)^2+2\sum x_k-(n+1)=\sum (1-x_k)^2+n-1+2a$
Let $x_0 \leq x_i$ for $i=1,...,n$
Case 1: $x_0 \leq a$
Then $\sum (1-x_k)^2 \geq (1-x_0)^2 \geq (1-a)^2$
Case 2: $x_0 \geq a$
Then $\sum (1-x_k)^2=\sum x_k(x_k-2+\frac{1}{x_k}) \geq a \sum (x_k-2+\frac{1}{x_k})=a(n+\frac{1}{a}-2(n+1)+n+\frac{1}{a})=(a-1)^2$
So $\sum x_k^2 \geq (a-1)^2+n-1+2a=n+a^2$
alinazarboland
24.04.2023 18:23
Lemma.let $x,y,z$ be three positive reals where $x+y+z=c_1 , \frac {1}{x} + \frac {1}{y} + \frac {1}{z} = c_2$ where $c_1,c_2$ are positive constants. Then $min{x^2+y^2+z^2}$ accurs when at least two of $x,y,z$ are equal. Proof. We have $\frac {1}{x} + \frac {1}{y} + \frac {1}{c_1-x-y} = c_2$.Now by lagrange multipliers theorem , the minimum of $\sum x^2$ accurs when : $$\frac {1}{(c_1-x-y)^2} - \frac {1}{x^2} = t.(2x+y-c_1)$$And similar for $y$ where $t$ is constant. Now by subtracting the equations for $x,y$ , and factoring $x-y$ out in the case where $x,y$ are inequal , we'll have $x^2y^2(x+y)=s$ where $s$ is constant.Note that in this case we're writing $z=c_1-x-y$ and the constant $t$ would be the same if we do this either for $y$ or $x$ too. Which means: $$x^2y^2(x+y)=x^2z^2(x+z)= z^2y^2(z+y)=s $$And again by subtracting these equations $x=y=z$ . So at least two of $x,y z$ are equal. Now in the main problem , among any three of variables two of them are equal(by fixing $n-2$ of them). So $n$ of them are equal and the rest is an easy bash using the fact $a<1$ and the minimum will accur at $(1,1,...,1,a)$ which gives us the bound $n+a^2$
MathSaiyan
08.06.2023 13:41
The idea is to take three variables and do smoothing. The key is the following claim:
Claim.
Suppose we want to maximize $a^2+b^2+c^2$ for positive reals $a,b,c$ such that $a+b+c = m$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = k$ where $m,k$ are constants, and $a\le b,c$. Then the minimum is achieved when $b=c$.
Proof.
Let $ab+bc+ca = p$, so $abc = \frac{p}{k}$. Consider the polynomial $P(x) = x^3 - mx^2 + px - \frac{p}{k}$, which by cardano has roots $a,b,c$. Observe that $a,b,c > \frac{1}{k}$, as
\[
\frac{1}{k} = \frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} < \frac{1}{\frac{1}{a}} < a
\]and the same for $b$ and $c$.
Since $a^2+b^2+c^2 = m^2-2p$, we want to maximize $p$. So, we want to find the greatest $p$ such that the cubic
\[
P(x)=x^3-mx^2+px-\frac{p}{k}
\]has three positive real roots. Observe that if we increase $p$, then $P(t)$ increases if $t>\frac{1}{k}$. So, as $a,b,c>\frac{1}{k}$, increasing $p$ will force the smallest root to move to the left, the middle root to move to the right, and the greatest root to move to the left. This is best seen by considering the graph of $P(x)$, or just noticing that $P$ is negative between the middle and greatest roots.
So, we'll keep increasing $p$ until the middle and greatest roots are the same, hence the result follows. Also notice the smallest root will decrease during this process, so we avoid problems with the smallest root becoming equal to the middle one. $\square$
Suppose $x_0 \le x_1 \le \dots \le x_n$. We are going to spam this claim setting $a=x_0$, $b = x_1$, $c=x_n$ (and reordering $x_1,x_2,\dots,x_n$ after each application). This process doesn't end, but it does converge, as $x_n-x_1$ will converge to $0$ after infinitely many applications. As $\sum x_i^2$ is continuous, this implies that we only need to prove the ineq in the $x_1=x_2 = \dots = x_n$ case.
(One can make this even more formal by considering the limits and operating a bit in a similar way to the proof of SMV theorem)
So we just need to cover this one case, and it's really easy to see that $x_0 = a$, $x_1=x_2 = \dots = x_n = 1$ gives the minimum, so the answer is $n+a^2$.
ChanandlerBong
18.06.2023 04:17
Here is a solution that I found. For any index $0\leq j\leq n$, notice that by Cauchy-Schwarz inequality, we have $$(n+a-x_{j})(n+\frac{1}{a}-\frac{1}{x_{j}})=\Sigma_{i\neq j} x_i \Sigma_{i\neq j}\frac{1}{x_i} \geq n^2$$which gives us $$(a+\frac{1}{a})n+2\geq (x_j+\frac{1}{x_j})n+\frac{x_{j}}{a}+\frac{a}{x_j}\geq (x_j+\frac{1}{x_j})n+2$$Therefore $a+\frac{1}{a}\geq x_j+\frac{1}{x_j}$, and because $a\in (0,1)$, we know that $a\leq x_j\leq \frac{1}{a}$ holds for $0\leq j\leq n$. Now, we have $${x_j}^{-1}(x_j-a)(x_j-1)^2\geq 0$$giving us the crucial inequality $${x_j}^2\geq (a+2)x_j+\frac{a}{x_j}-2a-1$$Summing it up over $0\leq j\leq n$ shows that $$\Sigma_{j=0}^{n}{x_j}^2\geq n+a^2$$equality happens if and only if one of $x_j$ equals to $a$ while the others equal to $1$.