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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
IZHO 2017 Functional equations
user01   51
N 5 minutes ago by lksb
Source: IZHO 2017 Day 1 Problem 2
Find all functions $f:R \rightarrow R$ such that $$(x+y^2)f(yf(x))=xyf(y^2+f(x))$$, where $x,y \in \mathbb{R}$
51 replies
+2 w
user01
Jan 14, 2017
lksb
5 minutes ago
chat gpt
fuv870   2
N 7 minutes ago by fuv870
The chat gpt alreadly knows how to solve the problem of IMO USAMO and AMC?
2 replies
fuv870
19 minutes ago
fuv870
7 minutes ago
Inequality with wx + xy + yz + zw = 1
Fermat -Euler   23
N 9 minutes ago by hgomamogh
Source: IMO ShortList 1990, Problem 24 (THA 2)
Let $ w, x, y, z$ are non-negative reals such that $ wx + xy + yz + zw = 1$.
Show that $ \frac {w^3}{x + y + z} + \frac {x^3}{w + y + z} + \frac {y^3}{w + x + z} + \frac {z^3}{w + x + y}\geq \frac {1}{3}$.
23 replies
Fermat -Euler
Nov 2, 2005
hgomamogh
9 minutes ago
Waiting for a dm saying me again "old geometry"
drmzjoseph   0
36 minutes ago
Source: Idk easy
Given $ABCD$ a tangencial quadrilateral that is not a rhombus, let $a,b,c,d$ be lengths of tangents from $A,B,C,D$ to the incircle of the quadrilateral which center is $I$. Let $M,N$ be the midpoints of $AC,BD$ resp. Prove that
\[ \frac{MI}{IN}=\frac{a+c}{b+d} \]
0 replies
drmzjoseph
36 minutes ago
0 replies
No more topics!
Simple inequality (86)
sqing   30
N Aug 28, 2020 by sqing
Source: Old or new
Let positive real numbers $a,b,c$ . Show that \[\sqrt{\frac{bc}{(c+a)(a+b)}}+\sqrt{\frac{ca}{(a+b)(b+c)}}+\sqrt{\frac{ab}{(b+c)(c+a)}}    >1 .\]
30 replies
sqing
Apr 23, 2013
sqing
Aug 28, 2020
Simple inequality (86)
G H J
Source: Old or new
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sqing
41023 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let positive real numbers $a,b,c$ . Show that \[\sqrt{\frac{bc}{(c+a)(a+b)}}+\sqrt{\frac{ca}{(a+b)(b+c)}}+\sqrt{\frac{ab}{(b+c)(c+a)}}    >1 .\]
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arqady
30149 posts
#2 • 3 Y
Y by sqing, Adventure10, Mango247
sqing wrote:
Let positive real numbers $a,b,c$ . Show that \[\sqrt{\frac{bc}{(c+a)(a+b)}}+\sqrt{\frac{ca}{(a+b)(b+c)}}+\sqrt{\frac{ab}{(b+c)(c+a)}}    >1 .\]
$\left(\sum_{cyc}\sqrt{\frac{bc}{(c+a)(a+b)}}\right)^2\sum_{cyc}b^2c^2(a+b)(a+c)\geq(ab+ac+bc)^3$.
Thus, it remains to prove that $(ab+ac+bc)^3>\sum_{cyc}b^2c^2(a^2+ab+ac+bc)$, which is obvious.
The following inequality is also true and also simple.
For non-negatives $a$, $b$ and $c$ such that $ab+ac+bc\neq0$ prove that:
\[\sum_{cyc}\sqrt{\frac{bc}{(c+a)(a+b)}} \geq\sqrt{1+\frac{10abc}{(a+b)(a+c)(b+c)}}\]
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sqing
41023 posts
#3 • 2 Y
Y by Adventure10, Mango247
Let $a,b,c\geq 0$, prove that
\[\sqrt{2}+\frac{4(3-2\sqrt{2})abc}{(a+b)(b+c)(c+a)}\geq \sum_{cyc}{\sqrt{\frac{ab}{(a+c)(b+c)}}}\geq 1+\dfrac{4abc}{(a+b)(b+c)(c+a)}\]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2836557#p2836557
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Sayan
2130 posts
#4 • 2 Y
Y by Adventure10, Mango247
Arqady's version is stronger,
we have:
For non-negatives $a$, $b$ and $c$ such that $ab+ac+bc\neq0$ prove that:
\[\sum_{cyc}\sqrt{\frac{bc}{(c+a)(a+b)}} \geq\sqrt{1+\frac{10abc}{(a+b)(a+c)(b+c)}}\geq 1+\dfrac{4abc}{(a+b)(b+c)(c+a)}\]
Z K Y
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sqing
41023 posts
#5 • 3 Y
Y by lminsl, Adventure10, Mango247
Thank Sayan.
For non-negatives $a$, $b$ and $c$ such that $ab+ac+bc\neq0$ prove that:\[\sqrt{\frac{bc}{(c+a)(a+b)}}+\sqrt{\frac{ca}{(a+b)(b+c)}}+\sqrt{\frac{ab}{(b+c)(c+a)}}  \geq 1\]
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teodora
6 posts
#6 • 2 Y
Y by Adventure10, Mango247
a solution for the inequality
Attachments:
Using inequality xy.zip (8kb)
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gxggs
172 posts
#7 • 2 Y
Y by Adventure10, Mango247
Sayan wrote:
Arqady's version is stronger,
we have:
For non-negatives $a$, $b$ and $c$ such that $ab+ac+bc\neq0$ prove that:
\[\sum_{cyc}\sqrt{\frac{bc}{(c+a)(a+b)}} \geq\sqrt{1+\frac{10abc}{(a+b)(a+c)(b+c)}}\geq 1+\dfrac{4abc}{(a+b)(b+c)(c+a)}\]
Here is the sharper one:
\[\sum_{cyc}{\sqrt{\frac{bc}{(a+b)(a+c)}}}\geq \sqrt{1+\sqrt{\frac{25abc}{2(a+b)(b+c)(c+a)}}}\]
Z K Y
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arqady
30149 posts
#8 • 2 Y
Y by sqing, Adventure10
For the proof of the last inequality we can use $\sum_{cyc}\sqrt{a(a+b)(a+c)}\geq2\sqrt{(a+b+c)(ab+ac+bc)}$.
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sqing
41023 posts
#9 • 2 Y
Y by Adventure10, Mango247
Let $a,b,c$ be positive real numbers . Show that \[\sqrt{\frac{bc}{(c+a)(a+b)}}+\sqrt{\frac{ca}{(a+b)(b+c)}}+\sqrt{\frac{ab}{(b+c)(c+a)}} \le \frac{3}{2} .\]
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Sayan
2130 posts
#10 • 1 Y
Y by Adventure10
sqing wrote:
Let $a,b,c$ be positive real numbers . Show that \[\sqrt{\frac{bc}{(c+a)(a+b)}}+\sqrt{\frac{ca}{(a+b)(b+c)}}+\sqrt{\frac{ab}{(b+c)(c+a)}} \le \frac{3}{2} .\]
We can use just AM-GM:
\begin{align*}\sum\sqrt{\frac{bc}{(c+a)(a+b)}} & \le \frac12\sum\left(\frac{c}{c+a}+\frac{b}{a+b}\right) \\ & =\frac12\sum\left(\frac{a}{a+b}+\frac{b}{a+b}\right)=\frac32\end{align*}
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sqing
41023 posts
#11 • 2 Y
Y by Adventure10, Mango247
Let $a,b,c$ be positive real numbers . Show that \[\sqrt{\frac{bc}{(c+2a)(2a+b)}}+\sqrt{\frac{ca}{(a+2b)(2b+c)}}+\sqrt{\frac{ab}{(b+2c)(2c+a)}} \geq1 .\]
Z K Y
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arqady
30149 posts
#12 • 1 Y
Y by Adventure10
sqing wrote:
Let $a,b,c$ be positive real numbers . Show that \[\sqrt{\frac{bc}{(c+2a)(2a+b)}}+\sqrt{\frac{ca}{(a+2b)(2b+c)}}+\sqrt{\frac{ab}{(b+2c)(2c+a)}} \geq1 \]
By Holder $\left(\sum_{cyc}\sqrt{\frac{ab}{(b+2c)(2c+a)}}\right)^2\sum_{cyc}a^2b^2(a+2c)(b+2c)\geq(ab+ac+bc)^3$.
Hence, it remains to prove that $(ab+ac+bc)^3\geq\sum_{cyc}a^2b^2(a+2c)(b+2c)$, which is $abc\sum_{cyc}c(a-b)^2\geq0$. Done!
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Wangzu
283 posts
#13 • 2 Y
Y by Adventure10, Mango247
For the folowing inequality:$$\sum_{cyc}\sqrt{\frac{bc}{(c+a)(a+b)}} \geq\sqrt{1+\frac{10abc}{(a+b)(a+c)(b+c)}}$$
By squaring the both sides we need to show that:

$$\sum \frac{ab}{(b+c)(c+a)}+2\sqrt{\frac{abc}{(a+b)(b+c)(c+a)}}\left(\sum \sqrt{\frac{a}{b+c}}\right)\geq 1+\frac{10abc}{(a+b)(b+c)(c+a)}$$If we use the well known inequality:

$$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\geq 2\sqrt{1+\frac{abc}{(a+b)(b+c)(c+a)}}$$
Then we need to prove that:

$$(a+b)(b+c)(c+a) \geq 8abc$$which is obvius by AM-GM
This post has been edited 4 times. Last edited by Wangzu, Oct 14, 2015, 4:39 PM
Reason: ok
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Mathskidd
605 posts
#14 • 2 Y
Y by Adventure10, Mango247
Dear Wangzu
Sorry, can you write fuller solution here?
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Wangzu
283 posts
#15 • 2 Y
Y by Adventure10, Mango247
gxggs wrote:
Sayan wrote:
Arqady's version is stronger,
we have:
For non-negatives $a$, $b$ and $c$ such that $ab+ac+bc\neq0$ prove that:
\[\sum_{cyc}\sqrt{\frac{bc}{(c+a)(a+b)}} \geq\sqrt{1+\frac{10abc}{(a+b)(a+c)(b+c)}}\geq 1+\dfrac{4abc}{(a+b)(b+c)(c+a)}\]
Here is the sharper one:
\[\sum_{cyc}{\sqrt{\frac{bc}{(a+b)(a+c)}}}\geq \sqrt{1+\sqrt{\frac{25abc}{2(a+b)(b+c)(c+a)}}}\]

By squaring the both sides, we need prove:

$$\sum \frac{ab}{(b+c)(c+a)}+2\sqrt{\frac{abc}{(a+b)(b+c)(c+a)}}\left(\sum \sqrt{\frac{a}{b+c}}\right)\geq 1+5\sqrt{\frac{abc}{2(a+b)(b+c)(c+a)}}$$
We still can use the inequality:

$$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\geq 2\sqrt{1+\frac{abc}{ (a+b)(b+c)(c+a)}}$$
Then let $x=\sqrt{\frac{abc}{(a+b)(b+c)(c+a)}}, x\leq \frac{1}{2\sqrt{2}}$ we got something obvious.

Or we can use the folowing:

$$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}+2(2\sqrt{2}-3)\sqrt{\frac{abc}{(a+b)(b+c)(c+a)}}\geq 2$$
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sqing
41023 posts
#16 • 2 Y
Y by Adventure10, Mango247
Let $a,b,c$ be positive real numbers . Show that \[\frac{a^3}{(c+2a)(2a+b)}+\frac{b^3}{(a+2b)(2b+c)}+\frac{c^3}{(b+2c)(2c+a)}\geq \frac{a+b+c}{9}.\]
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arqady
30149 posts
#17 • 2 Y
Y by Adventure10, Mango247
Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that:
\[\frac{1}{3}\sqrt{\frac{a^2+b^2+c^2}{3}}\leq\frac{a^3}{(c+2a)(2a+b)}+\frac{b^3}{(a+2b)(2b+c)}+\frac{c^3}{(b+2c)(2c+a)}\leq \frac{a^2+b^2+c^2}{3(a+b+c)}\]
This post has been edited 1 time. Last edited by arqady, Oct 16, 2015, 2:02 AM
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sqing
41023 posts
#18 • 2 Y
Y by Adventure10, Mango247
Let $a,b,c$ be non-negative real numbers such that $ab+ac+bc\neq0 $ . Show that \[\sqrt{\frac{bc}{(c+2a)(2a+b)}}+\sqrt{\frac{ca}{(a+2b)(2b+c)}}+\sqrt{\frac{ab}{(b+2c)(2c+a)}} \leq \sqrt{\frac{a^2+b^2+c^2}{ab+ac+bc}}.\]
This post has been edited 1 time. Last edited by sqing, Oct 16, 2015, 2:09 AM
Z K Y
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Wangzu
283 posts
#19 • 2 Y
Y by Adventure10, Mango247
sqing wrote:
Let $a,b,c$ be positive real numbers . Show that \[\frac{a^3}{(c+2a)(2a+b)}+\frac{b^3}{(a+2b)(2b+c)}+\frac{c^3}{(b+2c)(2c+a)}\geq \frac{a+b+c}{9}.\]

It's just AM-GM for: $\frac{a^3}{(c+2a)(2a+b)}+\frac{c+2a}{27}+\frac{2a+b}{27} \geq \frac{a}{3}$
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sqing
41023 posts
#20 • 2 Y
Y by Adventure10, Mango247
Wangzu wrote:
sqing wrote:
Let $a,b,c$ be positive real numbers . Show that \[\frac{a^3}{(c+2a)(2a+b)}+\frac{b^3}{(a+2b)(2b+c)}+\frac{c^3}{(b+2c)(2c+a)}\geq \frac{a+b+c}{9}.\]

It's just AM-GM for: $\frac{a^3}{(c+2a)(2a+b)}+\frac{c+2a}{27}+\frac{2a+b}{27} \geq \frac{a}{3}$
Yeah .
Thanks.
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arqady
30149 posts
#21 • 3 Y
Y by Monreal, Adventure10, Mango247
sqing wrote:
Let $a,b,c$ be non-negative real numbers such that $ab+ac+bc\neq0 $ . Show that \[\sqrt{\frac{bc}{(c+2a)(2a+b)}}+\sqrt{\frac{ca}{(a+2b)(2b+c)}}+\sqrt{\frac{ab}{(b+2c)(2c+a)}} \leq \sqrt{\frac{a^2+b^2+c^2}{ab+ac+bc}}.\]
By C-S $\left(\sum_{cyc}\sqrt{\frac{bc}{(c+2a)(2a+b)}}\right)^2\leq\sum_{cyc}\frac{c}{c+2a}\sum_{cyc}\frac{b}{b+2a}\leq \sqrt{\frac{a^2+b^2+c^2}{ab+ac+bc}}\cdot \sqrt{\frac{a^2+b^2+c^2}{ab+ac+bc}}=\frac{a^2+b^2+c^2}{ab+ac+bc}$.
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math90
1474 posts
#22 • 2 Y
Y by arqady, Adventure10
sqing wrote:
Let $a,b,c$ be positive real numbers . Show that \[\sqrt{\frac{bc}{(c+2a)(2a+b)}}+\sqrt{\frac{ca}{(a+2b)(2b+c)}}+\sqrt{\frac{ab}{(b+2c)(2c+a)}} \geq1 .\]
$\sum_{cyc}\sqrt{\frac{bc}{(2a+b)(2a+c)}}=\sum_{cyc}\frac{bc}{\sqrt{b(2a+c)}\sqrt{c(2a+b)}}\geq\sum_{cyc}\frac{2bc}{b(2a+c)+c(2a+b)}=1.$
This post has been edited 2 times. Last edited by math90, Oct 16, 2015, 7:35 AM
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sqing
41023 posts
#23 • 1 Y
Y by Adventure10
Very very nice .
Thank you very much .
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Monreal
96 posts
#24 • 1 Y
Y by Adventure10
arqady wrote:
sqing wrote:
Let $a,b,c$ be non-negative real numbers such that $ab+ac+bc\neq0 $ . Show that \[\sqrt{\frac{bc}{(c+2a)(2a+b)}}+\sqrt{\frac{ca}{(a+2b)(2b+c)}}+\sqrt{\frac{ab}{(b+2c)(2c+a)}} \leq \sqrt{\frac{a^2+b^2+c^2}{ab+ac+bc}}.\]
By C-S $\left(\sum_{cyc}\sqrt{\frac{bc}{(c+2a)(2a+b)}}\right)^2\leq\sum_{cyc}\frac{c}{c+2a}\sum_{cyc}\frac{b}{b+2a}\leq \sqrt{\frac{a^2+b^2+c^2}{ab+ac+bc}}\cdot \sqrt{\frac{a^2+b^2+c^2}{ab+ac+bc}}=\frac{a^2+b^2+c^2}{ab+ac+bc}$.

Do you have a nice proof for $\sum_{cyc}\frac{c}{c+2a} \leq \sqrt{\frac{a^2+b^2+c^2}{ab+ac+bc}}$. Thanks!
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arqady
30149 posts
#25 • 1 Y
Y by Adventure10
I have no a nice proof. $uvw$ kills it.
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luofangxiang
4613 posts
#26 • 2 Y
Y by Adventure10, Mango247
arqady wrote:
Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that:
\[\frac{1}{3}\sqrt{\frac{a^2+b^2+c^2}{3}}\leq\frac{a^3}{(c+2a)(2a+b)}+\frac{b^3}{(a+2b)(2b+c)}+\frac{c^3}{(b+2c)(2c+a)}\leq \frac{a^2+b^2+c^2}{3(a+b+c)}\]

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Mathskidd
605 posts
#27 • 2 Y
Y by Adventure10, Mango247
$\sqrt{\sum_{cyc}a^2\sum_{cyc}(a^2+bc)^2}\leq 6$
$\iff \sum_{cyc}(a^2+bc)^2\leq 12$
Can you elaborate the above inequality ?
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luofangxiang
4613 posts
#28 • 2 Y
Y by Adventure10, Mango247
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luofangxiang
4613 posts
#29 • 1 Y
Y by Adventure10
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SidVicious
584 posts
#31 • 2 Y
Y by Adventure10, Mango247
sqing wrote:
Let positive real numbers $a,b,c$ . Show that \[\sqrt{\frac{bc}{(c+a)(a+b)}}+\sqrt{\frac{ca}{(a+b)(b+c)}}+\sqrt{\frac{ab}{(b+c)(c+a)}}    >1 .\]

Let $x=\frac{a}{b+c}$, $y=\frac{b}{a+c}$, $z=\frac{c}{b+a}$. Then $xy+yz+xz+2xyz=1$ and we need to prove $\sqrt{xy}+\sqrt{yz}+\sqrt{zx}>1$. Now, let
$m=xy$, $n=yz$, $p=zx$. Then we need to prove $\sqrt{m}+\sqrt{n}+\sqrt{p}>1$ with $m+n+p+2\sqrt{mnp}=1$. We have that $\sqrt{m}+\sqrt{n}+\sqrt{p}>1$ is equivalent with $m+n+p+2(\sqrt{mn}+\sqrt{np}+\sqrt{pm})>1=m+n+p+2\sqrt{mnp}$. So we want to prove:
$\sqrt{mn}+\sqrt{np}+\sqrt{pm}>\sqrt{mnp}$. Note that $\sqrt{mn}+\sqrt{np}+\sqrt{pm} \geq 3*(mnp)^{\frac{1}{3}}$ so we want to prove $3*(mnp)^{\frac{1}{3}}>\sqrt{mnp}$ or $3^6*(mnp)^2>(mnp)^3$ or $mnp<3^6$ which is true because $m+n+p+2\sqrt{mnp}=1 \geq 3*(mnp)^{\frac{1}{3}}+2\sqrt{mnp}$. Q.E.D
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sqing
41023 posts
#32
Y by
sqing wrote:
Let $a,b,c$ be positive real numbers . Show that \[\sqrt{\frac{bc}{(c+a)(a+b)}}+\sqrt{\frac{ca}{(a+b)(b+c)}}+\sqrt{\frac{ab}{(b+c)(c+a)}} \le \frac{3}{2} .\]
Let positive real numbers $a,b,c$ . Show that
\[\sqrt{\frac{bc}{(c+a)(a+b)}}+\sqrt{\frac{ca}{(a+b)(b+c)}}+\sqrt{\frac{ab}{(b+c)(c+a)}} \leq\frac{3}{2}-\frac{(b-c)^2}{12(b+c)^2} .\]
sqing wrote:
Let $a,b,c$ be positive real numbers . Show that \[\sqrt{\frac{bc}{(c+2a)(2a+b)}}+\sqrt{\frac{ca}{(a+2b)(2b+c)}}+\sqrt{\frac{ab}{(b+2c)(2c+a)}} \geq1 .\]
N:
\[\sqrt {{\frac {bc}{ \left(c+2a\right)  \left( 2a+b\right) }}}\geq {
\frac {bc}{ab+bc+ca}}\]
This post has been edited 1 time. Last edited by sqing, Aug 28, 2020, 7:57 AM
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