Let $n$ be an integer number greater than $2$, let $x_{1},x_{2},\ldots ,x_{n}$ be $n$ positive real numbers such that \[\sum_{i=1}^{n}\frac{1}{x_{i}+1}=1\] and let $k$ be a real number greater than $1$. Show that: \[\sum_{i=1}^{n}\frac{1}{x_{i}^{k}+1}\ge\frac{n}{(n-1)^{k}+1}\] and determine the cases of equality.
Problem
Source: RMO
Tags: inequalities, inequalities proposed, Balkan
test20
10.04.2012 16:03
horizon wrote: Let $n$ be an integer number greater than $2$, let $x_{1},x_{2},\ldots ,x_{n}$ be $n$ positive real numbers such that \[\sum_{i=1}^{n}\frac{1}{x_{i}+1}=n\] and let $k$ be a real number greater than $1$. Show that: \[\sum_{i=1}^{n}\frac{1}{x_{i}^{k}+1}\ge\frac{n}{(n-1)^{k}+1}\] and determine the cases of equality. There are no positive real numbers $x_{1},x_{2},\ldots ,x_{n}$ that would satisfy the condition \[\sum_{i=1}^{n}\frac{1}{x_{i}+1}=n.\] I guess this condition should be changed to something like \[\sum_{i=1}^{n}\frac{1}{x_{i}+1}=1.\]
WakeUp
10.04.2012 17:03
I checked and you're right. It has now been changed..
mathocean97
07.12.2012 04:48
Sorry for reviving, but since nobody has posted a solution...
Lemma 1: $n^{k-1}\sum_{i=1}^{n}{a_i^k} \ge (\sum_{i=1}^{n}{a_i})^k$.
Proof: By Holder's Inequality,
$
\\
\\
(\sum_{i=1}^{n}{1^\frac{k}{k-1}})^\frac{k-1}{k}(\sum_{i=1}^{n}{a_i^k})^\frac{1}{k} \ge (\sum_{i=1}^{n}{a_i})$.
Now taking both sides to the $k^{th}$ power gives the result.
Now we perform the substitution $a_i = \frac{1}{x_i+1}$, which is equivalent to $x_i = \frac{1-a_i}{a_i}$.
So our condition becomes $\sum_{i=1}^n{a_i} = 1$.
Now substituting $x_i = \frac{1-a_i}{a_i}$ into the desired inequality transforms it to
$\sum_{i=1}^n{\frac{a_i^k}{a_i^k+(1-a_i)^k}} = \sum_{i=1}^n{\frac{a_i^k}{a_i^k+(\sum_{{j=1}_{j\not=i}}^n{a_j})^k}} \ge \frac{n}{(n-1)^k+1}$, since $\sum_{i=1}^n{a_i} = 1$.
But by Lemma 1 and Cauchy,
$
\\
\sum_{i=1}^n{\frac{a_i^k}{a_i^k+(\sum_{{j=1}_{j\not=i}}^n{a_j})^k}} \ge
\\
\sum_{i=1}^n{\frac{a_i^k}{a_i^k+((n-1)^{k-1}\sum_{{j=1}_{j\not=i}}^n{a_j^k})}
=
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\sum_{i=1}^n{\frac{a_i^{2k}}{a_i^{2k}+((n-1)^{k-1}\sum_{{j=1}_{j\not=i}}^n{a_i^ka_j^k})}}
\ge
\\
{\frac{(\sum_{i=1}^n{a_i^k})^2}{\sum_{i=1}^n{a_i^{2k}}+((n-1)^{k-1}\sum_{{i = 1},{j=1},{j\not=i}}^n{{a_i^ka_j^k})}}
$.
Now using the fact that $(n-1)\sum_{i=1}^n{a_i^{2k}} \ge \sum_{{i=1},{j=1},{j\not=i}}^n{a_i^ka_j^k}$,
${\frac{(\sum_{i=1}^n{a_i^k})^2}{\sum_{i=1}^n{a_i^{2k}}+((n-1)^{k-1}\sum_{{i = 1},{j=1},{j\not=i}}^n{{a_i^ka_j^k}}}}
\\
=
{\frac{\sum_{i=1}^n{a_i^{2k}}+\sum_{{i = 1},{j=1},{j\not=i}}^n{{a_i^ka_j^k}}}{\sum_{i=1}^n{a_i^{2k}}+((n-1)^{k-1}\sum_{{i = 1},{j=1},{j\not=i}}^n{{a_i^ka_j^k}})}}
\ge
\\
\frac{\frac{n}{(n-1)^k+1}({\sum_{i=1}^n{a_i^{2k}}+(n-1)^{k-1}\sum_{{i = 1},{j=1},{j\not=i}}^n{{a_i^ka_j^k}}})}{\sum_{i=1}^n{a_i^{2k}}+((n-1)^{k-1}\sum_{{i = 1},{j=1},{j\not=i}}^n{{a_i^ka_j^k}})} (*)
\\
=
\frac{n}{(n-1)^k+1}$.
Step (*) is true because $\frac{n}{(n-1)^k+1} < 1$, since $k > 1$.
sqing
07.04.2020 09:17
horizon wrote: Let $n$ be an integer number greater than $2$, let $x_{1},x_{2},\ldots ,x_{n}$ be $n$ positive real numbers such that $\sum_{i=1}^{n}\frac{1}{x_{i}+1}=1$ and let $k$ be a real number greater than $1$. Show that:\[\sum_{i=1}^{n}\frac{1}{x_{i}^{k}+1}\ge\frac{n}{(n-1)^{k}+1}\]and determine the cases of equality. Balkan MO Shortlist 2011 Let $a,b,c>0$ and $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1$ . For the integer $n\geq2$ ,prove that$$\frac{1}{1+a^n}+\frac{1}{1+b^n}+\frac{1}{1+c^n}\geq\frac{3}{1+2^n}$$
parola
26.05.2020 21:51
I am not sure if this is correct , as i saw this technique only once. Anyway, i think i managed to solve it with the main idea from a friend. Claim:$\sum x_i \geq n(n-1)$ Proof: As $f(x_i)=\frac{1}{x_i+1}$ is convex , we obtain by Jensen's inequality that $\frac{\sum f(x_i)}{n} \geq f(\frac{\sum x_i}{n})$.Let's note $\sum x_i =S$.By the condition we obtain that $\frac{1}{n} \geq \frac{n}{S+n}$ so the claim is proved. Note that by Holder's inequality $n^{k-1}\cdot \sum x_i^k \geq (\sum x_i)^k \geq n^k(n-1)^k$ Now, the main idea ( which I am not sure that i used it well, so please tell me if i am wrong) is to consider the generating series of the terms. The generating series of $\frac{1}{x_i^k+1}$ is :$\frac{1}{x_i^k+1}=1-x_i^k+x_i^{2k}-...$ and $\frac{n}{(n-1)^k+1}=n(1-(n-1)^k+(n-1)^{2k}-...$. Notice that the ones from LHS equlas n. If we prove that $\sum {x_i^{2xk}-x_i^{2xk-k}} \geq n((n-1)^{2xk}-(n-1)^{2xk-k}))$ , then applying it infinetly times we get the desired inequality. By Cebishev inequality $n\sum x_i^{2xk} \geq \sum x_i^{2xk-k} \sum x_i^k $ and by Holder's inequality again $n^{2xk-k-1} \sum x_i^{2xk-k} \geq(\sum x_i)^{2xk-k} \geq n^{2xk-k}(n-1)^{2xk-k}$ , so $\sum x_i^{2xk-k} \geq n(n-1)^{2xk-k}$. Now ,$\sum {x_i^{2xk}-x_i^{2xk-k}} \geq \sum x_i^{2xk-k}\cdot \frac{\sum x_i^k-n}{n} \geq n(n-1)^{2xk-k}\cdot \frac{n(n-1)^k-n}{n}=n((n-1)^{2xk}-(n-1)^{2xk-k}) $ finshing the proof . Hope it's ok.
mcmp
12.10.2024 09:25
Desmosable inequality. Let $y_i=\frac{1}{x_i+1}$ so that the condition collapses to $\sum_{i=1}^ny_i=1$ whilst the inequality reduces down to \[ \sum_{i=1}^n\frac{1}{\left(\frac{1}{y_i}-1\right)^k+1}\ge\frac{n}{(n-1)^k+1} \] We now claim that if $f_k(x)=\frac{1}{\left(\frac{1}{x}-1\right)^k+1}$ then: \[ f_k(x)\ge f_k’\left(\frac{1}{n}\right)\left(x-\frac{1}{n}\right)+f_k\left(\frac{1}{n}\right) \]with strict inequality everywhere except for $x=\frac{1}{n}$. This inequality can be checked using desmos and I am NOT bothered to do it by hand . Hence as $f_k\left(\frac{1}{n}\right)=\frac{1}{(n-1)^k+1}$: \begin{align*} \sum_{i=1}^nf_k(y_i)&\ge\sum_{i=1}^{n}f_k’\left(\frac{1}{n}\right)\left(y_i-\frac{1}{n}\right)+f_k\left(\frac{1}{n}\right)\\ &=f_k’\left(\frac{1}{n}\right)\left(-n\cdot\frac{1}{n}+\sum_{i=1}^ny_i\right)+\frac{n}{(n-1)^k+1}\\ &=\frac{n}{(n-1)^k+1} \end{align*}as desired! Equality when all $y_i=\frac{1}{n}$, i.e. $x_i=n-1$ for all $i$.