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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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0 replies
jlacosta
Mar 2, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Cool Number Theory
Fermat_Fanatic108   1
N 20 minutes ago by Fermat_Fanatic108
For an integer with 5 digits $n=abcde$ (where $a, b, c, d, e$ are the digits and $a\neq 0$) we define the \textit{permutation sum} as the value $$bcdea+cdeab+deabc+eabcd$$For example the permutation sum of 20253 is $$02532+25320+53202+32025=113079$$Let $m$ and $n$ be two fivedigit integers with the same permutation sum.
Prove that $m=n$.
1 reply
Fermat_Fanatic108
20 minutes ago
Fermat_Fanatic108
20 minutes ago
ratio chasing inside a triangle, segment trisecting
parmenides51   10
N 22 minutes ago by sangsidhya
Source: CRMO 2012 Region 2 p5
Let $ABC$ be a triangle. Let $D, E$ be a points on the segment $BC$ such that $BD =DE = EC$. Let $F$ be the mid-point of $AC$. Let $BF$ intersect $AD$ in $P$ and $AE$ in $Q$ respectively. Determine $BP:PQ$.
10 replies
parmenides51
Sep 30, 2018
sangsidhya
22 minutes ago
Geo: incircle, escircle, isotomic conjugate
XAN4   0
22 minutes ago
Source: Own
For $\triangle{ABC}$, Its incircle $\odot I$ and $A-$escircle $\odot I_A$ are tangent to $BC$ at $D$ and $E$ respectively. $AI$ intersects line $BC$ at $J$. Line $AD$ intersects $\odot I$ at $F$, and line $AE$ intersects $\odot I_A$ at $G$. Line $FG$ intersects $BC$ at $H$. Prove that $BJ=CH$.
0 replies
XAN4
22 minutes ago
0 replies
f(x)+f(x+y) \leq f(xy)+f(y)
augustin_p   7
N 25 minutes ago by MuradSafarli
Source: Estonia TST 2022
Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy the following condition for any real numbers $x{}$ and $y$ $$f(x)+f(x+y) \leq f(xy)+f(y).$$
7 replies
augustin_p
Jul 12, 2023
MuradSafarli
25 minutes ago
Integer equation in 3 variables
Kimchiks926   2
N 33 minutes ago by MuradSafarli
Source: Latvian TST for Baltic Way 2019 Problem 15
Determine all tuples of integers $(a,b,c)$ such that:
$$(a-b)^3(a+b)^2 = c^2 + 2(a-b) + 1$$
2 replies
Kimchiks926
May 29, 2020
MuradSafarli
33 minutes ago
Interesting inequality
sqing   2
N 36 minutes ago by SunnyEvan
Source: Own
Let $ a,b,c\geq 2  . $ Prove that
$$(a^2-1)(b-1)(c^2-1) -\frac{9}{4}abc\geq -9$$$$(a^2-1)(b-1)(c^2-1) -\frac{11}{5}abc\geq -\frac{43}{5}$$$$(a^2-1)(b-1)(c^2-1) -2abc\geq -7$$$$(a-1)(b^2-1)(c-1) -\frac{3}{4}abc\geq -3$$$$(a-1)(b^2-1)(c-1) -\frac{3}{5}abc\geq -\frac{9}{5}$$$$(a-1)(b^2-1)(c-1) -\frac{1}{2}abc\geq -1$$
2 replies
sqing
an hour ago
SunnyEvan
36 minutes ago
Interesting inequality
sqing   1
N 39 minutes ago by sqing
Source: Own
Let $ a,b,c\geq 2  . $ Prove that
$$(a^2-1)(b-\frac{3}{2})(c^2-1) - \frac{9}{4}abc\geq -15$$$$(a^2-1)(b-\frac{3}{2})(c^2-1) - 2abc\geq -\frac{73}{6}$$$$(a^2-1)(b-\frac{3}{2})(c^2-1) - abc\geq -\frac{7}{2}$$$$(a^2-2)(b-\frac{3}{2})(c^2-2) - abc\geq -6$$
1 reply
sqing
an hour ago
sqing
39 minutes ago
Hard problem involving circumcenter and concurrent lines
GeoMetrix   6
N 40 minutes ago by bin_sherlo
Source: AQGO 2020 Problem 3
Let $\triangle{ABC}$ be a triangle with circumcenter $O$. Let $M,N$ be the midpoints of $\overline{AB}$ and $\overline{AC}$ respectively and let $T$ be the projection of $O$ on $\overline{MN}$. Let $D$ be the projection of $A$ on $\overline{BC}$. Let $\overline{TD}$ intersect $\odot(BOC)$ at points $U$ and $V$. Let $\odot(AUV)$ intersct $\overline{MN}$ at points $X,Y$. Let $\overline{AY}$ intersect $\odot(AMN)$ at $R$ and $\overline{AX}$ intersect $\odot(AMN)$ at $S$. Then show that $\overline{AO},\overline{RS},\overline{MN}$ are concurrent.

Proposed by GeoMetrix
6 replies
GeoMetrix
Jun 20, 2020
bin_sherlo
40 minutes ago
Oi! These lines concur
Rg230403   17
N an hour ago by L13832
Source: LMAO 2021 P5, LMAOSL G3(simplified)
Let $I, O$ and $\Gamma$ respectively be the incentre, circumcentre and circumcircle of triangle $ABC$. Points $A_1, A_2$ are chosen on $\Gamma$, such that $AA_1 = AI = AA_2$, and point $A'$ is the foot of the altitude from $I$ to $A_1A_2$. If $B', C'$ are similarly defined, prove that lines $AA', BB'$ and $CC'$ concurr on $OI$.
Original Version from SL
Proposed by Mahavir Gandhi
17 replies
Rg230403
May 10, 2021
L13832
an hour ago
Find the period
Anto0110   1
N an hour ago by Anto0110
Let $a_1, a_2, ..., a_k, ...$ be a sequence that consists of an initial block of $p$ positive distinct integers that then repeat periodically. This means that $\{a_1, a_2, \dots, a_p\}$ are $p$ distinct positive integers and $a_{n+p}=a_n$ for every positive integer $n$. The terms of the sequence are not known and the goal is to find the period $p$. To do this, at each move it possible to reveal the value of a term of the sequence at your choice.
If $p$ is one of the first $k$ prime numbers, find for which values of $k$ there exist a strategy that allows to find $p$ revealing at most $8$ terms of the sequence.
1 reply
Anto0110
Yesterday at 7:37 PM
Anto0110
an hour ago
inequality
senku23   2
N an hour ago by User21837561
Let x,y,z in R+ prove that 8(x^3+y^3+z^3)2≥9(x^2+yz)(y^2+zx)(z^2+xy).
2 replies
senku23
3 hours ago
User21837561
an hour ago
Differentiable functional
bakerbakura   2
N an hour ago by Gryphos
Find all differentiable functions $ f;\mathbb{R}\to\mathbb{R}$ such that, for all real numbers $ a,b,t$ with $ 0<t<1$, $ t^2f(a)+(1-t^2)f(b)\geq f(ta+(1-t)b)$
2 replies
bakerbakura
Jan 11, 2010
Gryphos
an hour ago
function equation
cipher703516247   2
N an hour ago by luutrongphuc

Find all the functions $f: \mathbb R^{+} \to \mathbb R^{+}$such that:
$$f(xy^n +f(y)^{2n}) )=f(x)f(y)^n +(yf(y))^n $$
2 replies
cipher703516247
Feb 14, 2025
luutrongphuc
an hour ago
gcd (a^n+b,b^n+a) is constant
EthanWYX2009   78
N 2 hours ago by SimplisticFormulas
Source: 2024 IMO P2
Determine all pairs $(a,b)$ of positive integers for which there exist positive integers $g$ and $N$ such that
$$\gcd (a^n+b,b^n+a)=g$$holds for all integers $n\geqslant N.$ (Note that $\gcd(x, y)$ denotes the greatest common divisor of integers $x$ and $y.$)

Proposed by Valentio Iverson, Indonesia
78 replies
EthanWYX2009
Jul 16, 2024
SimplisticFormulas
2 hours ago
Happy Chinese New Year (2)
sqing   12
N Mar 28, 2021 by sqing
Source: Lijvzhi
Let $a,b,c$ are positive reals such that $ a\ge b+c. $ Prove that
$$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geq \frac{2}{3}(a+b+c)$$$$\left(\frac{a}{b+c}+\frac{b}{c+a}\right)\left(\frac{b}{c+a}+\frac{c}{a+b}\right)\left(\frac{c}{a+b}+\frac{a}{b+c}\right)
\geq \frac{32}{27} $$$$\frac{bc}{(c+a)(a+b)}+\frac{ca}{(a+b)(b+c)}+\frac{ ab}{(b+c)(c+a)} \geq \frac{7}{9}$$$$\sqrt{\frac{bc}{(c+a)(a+b)}}+\sqrt{\frac{ca}{(a+b)(b+c)}}+\sqrt{\frac{ ab}{(b+c)(c+a)}}\geq 1$$here
12 replies
sqing
Feb 12, 2021
sqing
Mar 28, 2021
Happy Chinese New Year (2)
G H J
Source: Lijvzhi
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sqing
41100 posts
#1 • 1 Y
Y by xyzz
Let $a,b,c$ are positive reals such that $ a\ge b+c. $ Prove that
$$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geq \frac{2}{3}(a+b+c)$$$$\left(\frac{a}{b+c}+\frac{b}{c+a}\right)\left(\frac{b}{c+a}+\frac{c}{a+b}\right)\left(\frac{c}{a+b}+\frac{a}{b+c}\right)
\geq \frac{32}{27} $$$$\frac{bc}{(c+a)(a+b)}+\frac{ca}{(a+b)(b+c)}+\frac{ ab}{(b+c)(c+a)} \geq \frac{7}{9}$$$$\sqrt{\frac{bc}{(c+a)(a+b)}}+\sqrt{\frac{ca}{(a+b)(b+c)}}+\sqrt{\frac{ ab}{(b+c)(c+a)}}\geq 1$$here
This post has been edited 1 time. Last edited by sqing, Feb 12, 2021, 1:08 PM
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Man0308Fong
23 posts
#2 • 1 Y
Y by MrOreoJuice
The first inequality
Proof:
$$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geq \frac{2}{3}(a+b+c)$$$$\Leftrightarrow \sum_{cyc}\frac{a^2}{(b+c)(a+b+c)}\geq\frac{2}{3}$$$$\Leftrightarrow \sum_{cyc}[\frac{a}{b+c}-\frac{a}{a+b+c}]\geq\frac{2}{3}$$$$\Leftrightarrow \sum_{cyc}\frac{a}{b+c}\geq\frac{5}{3}$$$$\Leftrightarrow 1+\frac{a-b-c}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{5}{3}$$$$\Leftrightarrow \frac{a-b-c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\geq\frac{8}{3}$$$$\Leftarrow \frac{a-b-c}{b+c}+\frac{4(a+b+c)}{2a+b+c}\geq\frac{8}{3}\hspace{3em} (\text{AM-HM inequality})$$$$\Leftrightarrow \frac{a-b-c}{b+c}+\frac{12(a+b+c)-8(2a+b+c)}{3(2a+b+c)} \geq 0$$$$\Leftrightarrow \frac{a-b-c}{b+c}-\frac{4a-4b-4c}{3(2a+b+c)}\geq 0$$$$\Leftrightarrow (a-b-c)[\frac{1}{b+c}-\frac{4}{3(2a+b+c)}]\geq 0$$$$\Leftrightarrow (a-b-c)\cdot\frac{6a-b-c}{3(b+c)(2a+b+c)}\geq 0$$$$\Leftrightarrow (a-b-c)\cdot[\frac{5a}{3(b+c)(2a+b+c)}+\frac{a-b-c}{3(b+c)(2a+b+c)}]\geq 0$$Equality occurs when $a=2b=2c$.
Q.E.D.
This post has been edited 1 time. Last edited by Man0308Fong, Feb 15, 2021, 2:37 AM
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sqing
41100 posts
#3 • 1 Y
Y by Mathskidd
Thanks.
sqing wrote:
Let $a,b,c$ are positive reals such that $ a\ge b+c. $ Prove that
$$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geq \frac{2}{3}(a+b+c)$$
Solution:
$$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geq \frac{2}{3}(a+b+c)$$$$\Leftrightarrow\frac{a^2}{b+c}+a+\frac{b^2}{c+a}+b+\frac{c^2}{a+b}+c \geq \frac{5}{3}(a+b+c)$$$$\Leftrightarrow \frac{a(a+b+c)}{b+c}+\frac{b(a+b+c)}{c+a}+\frac{c(a+b+c)}{a+b} \geq \frac{5}{3}(a+b+c)$$$$\Leftrightarrow \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}
\geq \frac{5}{3} $$Let $t=\frac{a}{b+c}$ $(t\ge 1)$
$$\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{{(b+c)}^2}{a(b+c)+2bc}\ge \frac{{(b+c)}^2}{a(b+c)+\frac{{(b+c)}^2}{2}}=\frac{2}{2t+1}$$$$ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geq t+\frac{2}{2t+1}$$It remains to prove
$$t+\frac{2}{2t+1}\ge \frac{5}{3} \iff \frac{(6t-1)(t-1)}{3}\ge 0$$which is true.
$$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geq \frac{2}{3}(a+b+c)$$Equality occurs when $a=2b=2c$.
This post has been edited 2 times. Last edited by sqing, Feb 28, 2021, 6:26 AM
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Man0308Fong
23 posts
#4 • 3 Y
Y by xyzz, MrOreoJuice, Mathskidd
The second inequality
Proof: Without loss of generality, let $a+b+c=1$, so $a\geq b+c \Leftrightarrow 1\geq a\geq\frac{1}{2}\geq b,c$
$$\left(\frac{a}{b+c}+\frac{b}{c+a}\right)\left(\frac{b}{c+a}+\frac{c}{a+b}\right)\left(\frac{c}{a+b}+\frac{a}{b+c}\right)\geq\frac{32}{27}$$$$\Leftrightarrow \prod_{cyc}\left(\frac{a}{1-a}+\frac{b}{1-b}\right)\geq\frac{32}{27}$$We use Jensen's inequality
$$\frac{b}{1-b}+\frac{c}{1-c}=\frac{1}{1-b}+\frac{1}{1-c}-2\geq\frac{2}{\frac{(1-b)+(1-c)}{2}}-2=\frac{2(1-a)}{1+a}$$Meanwhile, we use AM-GM inequality and Jensen's inequality again,
$$\left(\frac{a}{1-a}+\frac{b}{1-b}\right)\left(\frac{a}{1-a}+\frac{c}{1-c}\right)=\left(\frac{2a-1}{1-a}+\frac{1}{1-b}\right)\left(\frac{2a-1}{1-a}+\frac{1}{1-c}\right)$$$$=\left(\frac{2a-1}{1-a}\right)^2+\left(\frac{2a-1}{1-a}\right)\left(\frac{1}{1-b}+\frac{1}{1-c}\right)+\frac{1}{(1-b)(1-c)}$$$$\geq\left(\frac{2a-1}{1-a}\right)^2+\frac{2a-1}{1-a}\cdot\frac{4}{1+a}+\frac{4}{(1+a)^2}$$$$=\left(\frac{2a-1}{1-a}+\frac{2}{1+a}\right)^2$$Therefore, we got
$$\prod_{cyc}\left(\frac{a}{1-a}+\frac{b}{1-b}\right)\geq\left(\frac{2a-1}{1-a}+\frac{2}{1+a}\right)^2\cdot\frac{2(1-a)}{1+a}$$Denote $f(x)=\left(\frac{2x-1}{1-x}+\frac{2}{1+x}\right)^2\cdot\frac{2(1-x)}{1+x}$
$$f'(x)=2\left(\frac{2x-1}{1-x}+\frac{2}{1+x}\right)\cdot\frac{1}{(1+x)^2}\cdot\frac{x(5-3x)}{(1-x)(1+x)}$$When $x\in[\frac{1}{2},1]$, $\left(\frac{2x-1}{1-x}+\frac{2}{1+x}\right)>0$, $\frac{1}{(1+x)^2}>0$, $\frac{x(5-3x)}{(1-x)(1+x)}>0$
Therefore $f'(x)>0$.
In other word, $f(a) \geq f(\frac{1}{2}) = \left(\frac{4}{3}\right)^2\times\frac{2}{3}=\frac{32}{27}$
$$\left(\frac{a}{b+c}+\frac{b}{c+a}\right)\left(\frac{b}{c+a}+\frac{c}{a+b}\right)\left(\frac{c}{a+b}+\frac{a}{b+c}\right)\geq f(a) \geq f(\frac{1}{2})=\frac{32}{27}$$Equality holds when $a=2b=2c$.
Q.E.D.
This post has been edited 1 time. Last edited by Man0308Fong, Feb 15, 2021, 2:38 AM
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Man0308Fong
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#5 • 2 Y
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The third inequality
Proof: Denote $k=a-b-c \geq 0$
$$\frac{bc}{(c+a)(a+b)}+\frac{ca}{(a+b)(b+c)}+\frac{ ab}{(b+c)(c+a)} \geq \frac{7}{9}$$$$\Leftrightarrow 9\sum_{cyc}bc(b+c)-7(a+b)(b+c)(c+a)\geq 0$$$$\Leftrightarrow 9[bc(b+c)+c(b+c+k)(b+2c+k)+(b+c+k)b(2b+c+k)]-7(b+c)(b+2c+k)(2b+c+k)\geq 0$$$$\Leftrightarrow 4b^3+4c^3-4b^2c-4bc^2+2k^2(b+c)+6k(b^2-bc+c^2)\geq 0$$$$\Leftrightarrow 2(b+c)(b-c)^2+k^2(b+c)+3k(b^2-bc+c^2)\geq 0$$Equality holds when $k=0$ and $b=c$, i.e. $a=2b=2c$.
Q.E.D
This post has been edited 1 time. Last edited by Man0308Fong, Feb 14, 2021, 3:19 PM
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sqing
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#6
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Thanks.
Let $a, b$ and $c$ be non-negative numbers such that $a \geq b+c.$ Prove that
$$\frac{a}{b+c} + \frac{b}{c+a} +\frac{2c}{a+b} \geq    \frac{4\sqrt 2}{3}  $$
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#7 • 1 Y
Y by Yeetopia
It's technically not "Chinese" New Year. The correct term is "Lunar" New Year. A lot of countries follow the lunar calendar, not just China. So TECHNICALLY it's not Chinese New Year.
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#8
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Let $a, b$ and $c$ be non-negative numbers such that $a \geq b+c.$ Prove that
$$\frac{a}{b+c} + \frac{b}{c+a} +\frac{2c}{a+b} \geq    \frac{4\sqrt 2}{3}  $$Equality holds when $ a=\frac{4+3\sqrt 2}{2}c, b=\frac{2+3\sqrt 2}{2}c.$
Let $ a,b,c>0 $ and $ a\geq b+c.$ Prove that
$$\frac{a}{b+c} \left( \frac{b}{c+a}+\frac{c}{a+b} \right) \geq \frac{b}{2c+b} + \frac{c}{c+2b}\geq\frac{2}{3}$$$$ \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \geq 1+\frac{b}{2c+b} + \frac{c}{c+2b}$$h
Thanks. Just a title or symbol
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Man0308Fong
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#9
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The fourth inequality
Proof:
$$\sqrt{\frac{bc}{(c+a)(a+b)}}+\sqrt{\frac{ca}{(a+b)(b+c)}}+\sqrt{\frac{ ab}{(b+c)(c+a)}}\geq 1$$$$\Leftrightarrow \sum_{cyc}\sqrt{bc(b+c)}\geq\sqrt{(a+b)(b+c)(c+a)}$$$$\Leftrightarrow \sum_{cyc}bc(b+c)+2\sqrt{abc}\sqrt{c(b+c)(c+a)}\geq(a+b)(b+c)(c+a)=\sum_{sym}a^2b+2abc$$$$\Leftrightarrow \sum_{sym}b^2c+2\sqrt{abc}\sqrt{c(b+c)(c+a)}\geq(a+b)(b+c)(c+a)=\sum_{sym}a^2b+2abc$$$$\Leftrightarrow \sum_{cyc}\sqrt{c(b+c)(c+a)}\geq\sqrt{abc}$$In the meantime,
$$\sum_{cyc}\sqrt{c(b+c)(c+a)}\geq\sum_{cyc}{c\cdot 2\sqrt{bc} \cdot 2\sqrt{ca}}=2\sum_{cyc}c\sqrt[4]{ab}\geq 6\sqrt[3]{c\sqrt[4]{ab}\cdot a\sqrt[4]{bc}\cdot b\sqrt[4]{ca}}=6\sqrt{abc}>\sqrt{abc}$$Equality holds when $abc=0$, i.e. equality can't occur.
Q.E.D.

##The inequality should be $\forall a,b,c \in \mathbb{R}^{+}$, we got
$$\sqrt{\frac{bc}{(c+a)(a+b)}}+\sqrt{\frac{ca}{(a+b)(b+c)}}+\sqrt{\frac{ ab}{(b+c)(c+a)}}>1$$
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sqing
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#10
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Thanks.
Show that for $a,b,c > 0$ the following inequality holds:
$$\frac{\sqrt{ab}}{a+b+2c}+\frac{\sqrt{bc}}{b+c+2a}+\frac{\sqrt{ca}}{c+a+2b} \le \frac {3}{4}$$Let $a,b,c$ be positive real numbers. Prove that $$\frac{ab}{a+b+2c}+\frac{bc}{b+c+2a}+\frac{ca}{c+a+2b}\leq \frac{a+b+c}{4}$$Let $a$, $b$, $c$ be positive reals such that $a+b+c=3$. Show that$$\frac{ab}{2a+b+c} + \frac{bc}{2b+c+a}+ \frac{ca}{2c+a+b}\leq \frac{3}{4}$$Very easy.
$$\implies$$$$\sqrt{\frac{ab}{2a+b+c}} + \sqrt{\frac{bc}{2b+c+a}} + \sqrt{\frac{ca}{2c+a+b}} \leq \frac{3}{2}$$(Crux (3)2021,4624)
Let $a$, $b$, $c$ be positive reals such that $a+b+c=3$. Show that$$\frac{ab}{ka+b+c} + \frac{bc}{kb+c+a}+ \frac{ca}{kc+a+b}\leq \frac{3}{k+2}$$Where $k\geq1 .$
$$\implies$$Let $a$, $b$, $c$ be positive reals such that $a+b+c=3$. Show that
$$\sqrt{\frac{ab}{ka+b+c}} + \sqrt{\frac{bc}{kb+c+a}} + \sqrt{\frac{ca}{kc+a+b}} \leq \frac{3}{\sqrt{k+2}}$$Where $k\geq1 .$
Solution of Zhangyanzong:
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#12
Y by
sqing wrote:
Thanks.
sqing wrote:
Let $a,b,c$ are positive reals such that $ a\ge b+c. $ Prove that
$$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geq \frac{2}{3}(a+b+c)$$
Solution:
$$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geq \frac{2}{3}(a+b+c)$$$$\Leftrightarrow\frac{a^2}{b+c}+a+\frac{b^2}{c+a}+b+\frac{c^2}{a+b}+c \geq \frac{5}{3}(a+b+c)$$$$\Leftrightarrow \frac{a(a+b+c)}{b+c}+\frac{b(a+b+c)}{c+a}+\frac{c(a+b+c)}{a+b} \geq \frac{5}{3}(a+b+c)$$$$\Leftrightarrow \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}
\geq \frac{5}{3} $$Let $t=\frac{a}{b+c}$ $(t\ge 1)$
$$\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{{(b+c)}^2}{a(b+c)+2bc}\ge \frac{{(b+c)}^2}{a(b+c)+\frac{{(b+c)}^2}{2}}=\frac{2}{2t+1}$$$$ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geq t+\frac{2}{2t+1}$$It remains to prove
$$t+\frac{2}{2t+1}\ge \frac{5}{3} \iff \frac{(6t-1)(t-1)}{3}\ge 0$$which is true.
$$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geq \frac{2}{3}(a+b+c)$$Equality occurs when $a=2b=2c$.


Dear Sir Sqing ,
When did you complete the proof?
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#13
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Let $a,b,c$ are positive reals such that $ a\ge b+c. $ Prove that
$$ \left ( a+b+c\right )\left (\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )\geq 10$$$$ \left ( a^2+b^2+c^2\right )\left (\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \right )\geq \frac{27}{2}$$$$ \left ( a^3+b^3+c^3\right )\left (\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3} \right )\geq \frac{85}{4}$$$$ \left ( a^4+b^4+c^4\right )\left (\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4} \right )\geq \frac{297}{8}$$
sqing wrote:
Let $a,b,c$ are positive reals such that $ a\ge b+c. $ Prove that
$$1>\frac{bc}{(c+a)(a+b)}+\frac{ca}{(a+b)(b+c)}+\frac{ ab}{(b+c)(c+a)} \geq \frac{7}{9}$$
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Let $a,b,c$ are non-negative numbers such that $ a\ge 2(b+c). $ Prove that$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq 2+\frac{a^2+b^2+c^2}{5(ab+bc+ca)} $$here
sqing wrote:
Let $a,b,c$ are positive reals such that $ a\ge b+c. $ 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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Let $a,b,c$ be positive numbers. Probe that
$$\sqrt{\frac{bc}{(c+a)(a+b)}}+\sqrt{\frac{ca}{(a+b)(b+c)}}+\sqrt{\frac{ ab}{(b+c)(c+a)}} \geq \frac{9 a b c(a+b+c)}{2(a b+b c+c a)^{2}} $$
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