Generalization of Janous'one

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Generalization of Janous'one

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Source: Zhanyanzong

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sqing

#1 h Nov 3, 2020, 6:45 AM

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Let $a_1,a_2,\cdots,a_n (n\ge 2)$ be positive numbers such that $a_1\geq a_2+a_3+\cdots+a_k$ $(k\geq 2).$ Prove that $\left(a_1+a_2+\cdots+a_n\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right)\geq n^2+(k-2)^2.$

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#2 h Nov 3, 2020, 6:58 AM

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sqing wrote:

Let $a_1,a_2,\cdots,a_n (n\ge 2)$ be positive numbers such that $a_1\geq a_2+a_3+\cdots+a_k$ $(k\geq 2).$ Prove that $\left(a_1+a_2+\cdots+a_n\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right)\geq n^2+(k-2)^2.$

Sorry, but $k$ and $n$ depend to each other?

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sqing

#3 h Nov 3, 2020, 7:57 AM

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Auditor wrote:

sqing wrote:

Let $a_1,a_2,\cdots,a_n (n\ge 2)$ be positive numbers such that $a_1\geq a_2+a_3+\cdots+a_k$ $(k\geq 2).$ Prove that $\left(a_1+a_2+\cdots+a_n\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right)\geq n^2+(k-2)^2.$

Sorry, but $k$ and $n$ depend to each other?

Yeah.
Therefore , it is very nice.

Let $a,b,c$ and $d$ be positive real numbers such that $a \geq b+c.$ Proof that $\left(a+b+c+d\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\geq 17 .$ Let $a,b,c$ and $d$ be positive real numbers such that $a \geq b+c+d.$ Proof that $\left(a+b+c+d\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\geq 20 .$

This post has been edited 1 time. Last edited by sqing, Nov 3, 2020, 8:04 AM

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sqing

#5 h Nov 3, 2020, 9:23 AM

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themathematicalmantra wrote:

yes i find this inequality quite fascinating. Came across such elegant problem in a while. Will solve and send solution

Thank you very much.

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#6 h Nov 3, 2020, 9:40 AM

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Just use Cauchy-Schwarz:
$((a_1+a_2+\ldots+a_k)+(a_{k+1}+\ldots+a_n))\left(\left(\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_k}\right)+\frac{1}{a_{k+1}}+\ldots+\frac{1}{a_n}\right)\geq\left(\sqrt{(a_1+a_2+\ldots+a_k)\left(\frac{1}{a_1}+\frac{1}{a_2}\ldots+\frac{1}{a_k}\right)}+n-k\right)^2.$ Now, denote $s=a_2+\ldots+a_k$ and $a=a_1$ . Again, due to Cauchy-Schwarz (or AM-HM) we have
$\frac{1}{a_1}+\frac{1}{a_2}\ldots+\frac{1}{a_k}\geq\frac{1}{a}+\frac{(k-1)^2}{s}.$ Thus, it remains to prove that
$\left(\sqrt{(a+s)\left(\frac{1}{a}+\frac{(k-1)^2}{s}\right)}+n-k\right)^2\geq n^2+(k-2)^2.$ Consider the expression
$(a+s)\left(\frac{1}{a}+\frac{(k-1)^2}{s}\right)=(1+(k-1)^2)+\frac{s}{a}+(k-1)^2\cdot\frac{a}{s}.$ Since $a\geq s$ and function $f(t)=\frac{1}{t}+(k-1)^2t$ is increasing on $[1+\infty)$ (due to $k\geq 2$ ) we have
$\left(\sqrt{(a+s)\left(\frac{1}{a}+\frac{(k-1)^2}{s}\right)}+n-k\right)^2\geq \left(\sqrt{2(k-1)^2+2}+n-k\right)^2.$ Therefore it suffices to prove that
$\left(\sqrt{2(k-1)^2+2}+n-k\right)^2\geq n^2+(k-2)^2$ or
$2n\cdot\left(\sqrt{2(k-1)^2+2}-k\right)+\left(\sqrt{2(k-1)^2+2}-k\right)^2\geq (k-2)^2,$ so since $\sqrt{2(k-1)^2+2}-k\geq 0$ we just need to check this inequality when $n=k$ (and it's easier to check initial inequality) or
$2(k-1)^2+2\geq k^2+(k-2)^2,$ which is actually an equality.

Note. This solution isn't really artificial and is quite straightforward. Some motivation: firstly, we don't really care about $a_{k+1},\ldots,a_n$ , so we can optimize the left hand for variables $a_{k+1},\ldots,a_n$ (so the first inequality makes all $a_{k+1},\ldots,a_n$ equal to 1). Secondly, we have restriction only on the sum $a_1+\ldots+a_k$ , so we try to optimize left hand for $a_2,\ldots,a_k$ while the sum $a_2+\ldots+a_k$ is fixes (so our second inequality makes all variables $a_{2},\ldots,a_{k+1}$ equal). Basically, at this point we've reduced our problem to the case $a_{k+1}=\ldots=a_n=1$ and $a_2=\ldots=a_{k}=\frac{s}{k-1}$ with the only resrtriction $a_1\geq s$ . Finally, the left hand now is a function of $t=a/s$ , so we can find it's minimum on $[1+\infty)$ which is attained at $t=1$ . After that it remains to check simple inequality of variables $n$ and $k$ (and again we can apply our "optimizing principle" since $n\geq k$ and $LHS-RHS$ is an increasing function of $n$ ).

This post has been edited 2 times. Last edited by richrow12, Nov 3, 2020, 1:48 PM

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#7 h Nov 3, 2020, 9:49 AM

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Solution

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sqing

#8 h Nov 3, 2020, 10:28 AM

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sqing wrote:

Let $a,b,c$ and $d$ be positive real numbers such that $a \geq b+c.$ Proof that $\left(a+b+c+d\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\geq 17 .$

Let $a,b,c$ and $d$ be positive real numbers such that $a \geq b+c+d.$ Proof that $\left(a+b+c+d\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\geq 17 .$
Thank you very much.

This post has been edited 1 time. Last edited by sqing, Jul 20, 2022, 2:40 PM

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#9 h Nov 3, 2020, 12:19 PM

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richrow12 wrote:

Just use Cauchy-Schwarz:
$((a_1+a_2+\ldots+a_k)+(a_{k+1}+\ldots+a_n))\left(\left(\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_k}\right)+\frac{1}{a_{k+1}}+\ldots+\frac{1}{a_n}\right)\geq\left(\sqrt{(a_1+a_2+\ldots+a_k)\left(\frac{1}{a_1}+\frac{1}{a_2}\ldots+\frac{1}{a_k}\right)}+n-k\right)^2.$ Now, denote $s=a_2+\ldots+a_k$ and $a=a_1$ . Again, due to Cauchy-Schwarz (or AM-HM) we have
$\frac{1}{a_1}+\frac{1}{a_2}\ldots+\frac{1}{a_k}\geq\frac{1}{a}+\frac{(k-1)^2}{s}.$ Thus, it remains to prove that
$\left(\sqrt{(a+s)\left(\frac{1}{a}+\frac{(k-1)^2}{s}\right)}+n-k\right)^2\geq n^2+(k-2)^2.$ Consider the expression
$(a+s)\left(\frac{1}{a}+\frac{(k-1)^2}{s}\right)=(1+(k-1)^2)+\frac{s}{a}+(k-1)^2\cdot\frac{a}{s}.$ Since $a\geq s$ and function $f(t)=\frac{1}{t}+(k-1)^2t$ is increasing on $[1+\infty)$ (due to $k\geq 2$ ) we have
$\left(\sqrt{(a+s)\left(\frac{1}{a}+\frac{(k-1)^2}{s}\right)}+n-k\right)^2\geq \left(\sqrt{2(k-1)^2+2}+n-k\right)^2.$ Therefore it suffices to prove that
$\left(\sqrt{2(k-1)^2+2}+n-k\right)^2\geq n^2+(k-2)^2$ or
$2n\cdot\left(\sqrt{2(k-1)^2+2}-k\right)+\left(\sqrt{2(k-1)^2+2}-k\right)^2\geq (k-2)^2,$ so since $\sqrt{2(k-1)^2+2}-k\geq 0$ we just need to check this inequality when $n=k$ (and it's easier to check initial inequality) or
$2(k-1)^2+2\geq k^2+(k-2)^2,$ which is actually an equality.

Note. This solution isn't really artificial and is quite straightforward. Some motivation: firstly, we don't really care about $a_{k+1},\ldots,a_n$ , so we can optimize the left hand for variables $a_{k+1},\ldots,a_n$ (so the first inequality makes all $a_{k+1},\ldots,a_n$ equal to 1). Secondly, we have restriction only on the sum $a_1+\ldots+a_k$ , so we try to optimize left hand for $a_2,\ldots,a_k$ while the sum $a_2+\ldots+a_k$ is fixes (so our second inequality makes all variables $a_{2},\ldots,a_{k+1}$ equal). Basically, at this point we've reduced our problem to the case $a_{k+1}=\ldots=a_n=1$ and $a_2=\ldots=a_{k}=\frac{s}{k-1}$ with the only resrtriction $a_1\geq s$ . Finally, the left hand now is a function of $t=a/s$ , so we can find it's minimum on $[1+\infty)$ which attains at $t=1$ . After that it remains to check simple inequality of variables $n$ and $k$ (and again we can apply our "optimizing principle" since $n\geq k$ and $LHS-RHS$ is an increasing function of $n$ ).

But we don't know whether $n \ge k$ or not? Let me know if I am wrong.

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#10 h Nov 3, 2020, 12:37 PM

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@above We are only given $a_1, \dots, a_n$ . If $k > n$ , then there's a variable $a_k$ which is not given.

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yds

#11 h Nov 7, 2020, 11:25 AM

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sqing wrote:

Let $a_1,a_2,\cdots,a_n (n\ge 2)$ be positive numbers such that $a_1\geq a_2+a_3+\cdots+a_k$ $(k\geq 2).$ Prove that $\left(a_1+a_2+\cdots+a_n\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right)\geq n^2+(k-2)^2.$

$\text{Dear sqing can you add"n variable"or "Multivariate inequality",you know we AOPS' retrieval function is poor}$
$\text{Thank you very much!}$

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sqing

#12 h Nov 9, 2020, 12:29 AM

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Let $a_1,a_2,\cdots,a_n (n\ge 2)$ be positive numbers such that $a_1\geq a_2+a_3+\cdots+a_k$ $(n\geq k\geq 2).$ Prove that $\left(a^m_1+a^m_2+\cdots+a^m_n\right)\left(\frac{1}{a^m_1}+\frac{1}{a^m_2}+\cdots+\frac{1}{a^m_n}\right)\geq n^2+\frac{\left((k-1)^{|m|}-1\right)^2}{(k-1)^{|m|-1}}.$ Where $m\in R.$

This post has been edited 3 times. Last edited by sqing, Nov 9, 2020, 5:27 AM

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