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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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0 replies
jlacosta
May 1, 2025
0 replies
Orthocenter lies on circumcircle
whatshisbucket   89
N 26 minutes ago by Mathandski
Source: 2017 ELMO #2
Let $ABC$ be a triangle with orthocenter $H,$ and let $M$ be the midpoint of $\overline{BC}.$ Suppose that $P$ and $Q$ are distinct points on the circle with diameter $\overline{AH},$ different from $A,$ such that $M$ lies on line $PQ.$ Prove that the orthocenter of $\triangle APQ$ lies on the circumcircle of $\triangle ABC.$

Proposed by Michael Ren
89 replies
whatshisbucket
Jun 26, 2017
Mathandski
26 minutes ago
Hard math inequality
noneofyou34   5
N 34 minutes ago by JARP091
If a,b,c are positive real numbers, such that a+b+c=1. Prove that:
(b+c)(a+c)/(a+b)+ (b+a)(a+c)/(c+b)+(b+c)(a+b)/(a+c)>= Sqrt.(6(a(a+c)+b(a+b)+c(b+c)) +3
5 replies
noneofyou34
Sunday at 2:00 PM
JARP091
34 minutes ago
Interesting inequalities
sqing   0
37 minutes ago
Source: Own
Let $ a,b>0 $. Prove that
$$\frac{ab-1} {ab(a+b+2)} \leq \frac{1} {8}$$$$\frac{2ab-1} {ab(a+b+1)} \leq 6\sqrt 3-10$$
0 replies
sqing
37 minutes ago
0 replies
Inspired by SXJX (12)2022 Q1167
sqing   1
N an hour ago by sqing
Source: Own
Let $ a,b,c>0 $. Prove that$$\frac{kabc-1} {abc(a+b+c+8(2k-1))}\leq \frac{1}{16 }$$Where $ k>\frac{1}{2}.$
1 reply
sqing
Yesterday at 4:01 AM
sqing
an hour ago
No more topics!
Old problem
chien than   46
N Aug 15, 2024 by sqing
Source: MOSP 2001
Let $ a;b;c$ be positive real numbers satisfying $ abc=1$
Prove that:
$ (a+b)(b+c)(c+a) \geq 4(a+b+c-1)$
46 replies
chien than
Sep 30, 2007
sqing
Aug 15, 2024
Old problem
G H J
G H BBookmark kLocked kLocked NReply
Source: MOSP 2001
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SBM
780 posts
#35
Y by
sqing wrote:
Ji Chen (2019-09-21):
Let $ a,b,c$ be positive real numbers .Prove that$$c(a^3+b^3+c^3-3-4c)> 7(abc-ab-c).$$(p/6262708495)

We have:
$$c(a^3+b^3+c^3-3-4c)-7(abc-ab-c)$$$$= -\left[ 3ac+3bc+7c-7 \right] ab+c \left[  \left( a
+b \right) ^{3}+\left({c}^{3}-4c+4\right) \right]$$\[ = \left[ {\frac{1}{4}{\mkern 1mu} {{\left( {a - b} \right)}^2} + \frac{{{{\left( {3{\mkern 1mu} c\left( {a + b} \right) - 14{\mkern 1mu} c + 14} \right)}^2}}}{{108{\mkern 1mu} {c^2}}}} \right]\left( {3{\mkern 1mu} ac + 3{\mkern 1mu} bc + 7{\mkern 1mu} c - 7} \right) + \frac{{27{\mkern 1mu} {c^6} - 108{\mkern 1mu} {c^4} - 235{\mkern 1mu} {c^3} + 1029{\mkern 1mu} {c^2} - 1029{\mkern 1mu} c + 343}}{{27{c^2}}}\]Now$,$ easy prove.
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sqing
42260 posts
#36
Y by
Thanks.
Let $ a,b,c$ be positive real numbers satisfying $a^2+b^2+c^2=3.$ Prove that$$4(a+b+c-2)< (a+b)(b+c)(c+a) \leq 4(a+b+c-1)$$
This post has been edited 1 time. Last edited by sqing, Aug 8, 2021, 7:10 AM
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sqing
42260 posts
#38
Y by
Let $ a,b,c$ be positive real numbers satisfying $ abc=1.$ Prove that
$$ (a+b)(b+c)(c+a) \geq 8(a+b+c-2)$$Let $ a,b,c$ be non-negative numbers satisfying $ a(b+1)c=1.$ Prove that
$$ (a+b)(b+c)(c+a) \geq 2(a+b+c-1)$$h
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ZJ42
8166 posts
#40
Y by
why is this thread still active from 2007? and more importantly why does sqing bump it every few years?
This post has been edited 1 time. Last edited by ZJ42, Oct 22, 2021, 4:45 PM
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Jwenslawski
4344 posts
#41
Y by
Why does sqing post a bunch of problems though on other peoples threads?
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franzliszt
23531 posts
#42 • 1 Y
Y by Jwenslawski
I believe it is because sqing posts problems related to the original, so that people can try to solve them.
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Jwenslawski
4344 posts
#43
Y by
franzliszt wrote:
I believe it is because sqing posts problems related to the original, so that people can try to solve them.

But why doesn't he make a thread of his own problems?
Z K Y
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franzliszt
23531 posts
#44 • 1 Y
Y by Jwenslawski
Because then there would be too many threads :P
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jasperE3
11354 posts
#45 • 1 Y
Y by megarnie
You can get postbanned for creating too many threads.
Z K Y
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Kunihiko_Chikaya
14514 posts
#46
Y by
chien than wrote:
Let $ a;b;c$ be positive real numbers satisfying $ abc=1$
Prove that:
$ (a+b)(b+c)(c+a) \geq 4(a+b+c-1)$

$\left ( a+b \right )\left ( b+c \right )\left ( c+a \right )\geq 4\left ( a+b+c-1 \right ) $

$\Longleftrightarrow (a+b+c)(ab+bc+ca)-abc\geq 4\left ( a+b+c-1 \right )$

$\Longleftrightarrow (a+b+c)(ab+bc+ca)+3\geq 4(a+b+c).$

Proof :

$ab+bc+ca+\frac{3}{a+b+c}\geq \sqrt{3abc(a+b+c)}+\frac{3}{a+b+c}$

$=\sqrt{3}x+\frac{3}{x^2}$ where $x=\sqrt{a+b+c}\geq \sqrt{3}$ by AM-GM

$=\frac{(x-\sqrt{3})\{(x-\sqrt{3})(\sqrt{3}x+2)+\sqrt{3}\}}{x^2}+4\geq 4.$

Equality : $x = \sqrt{3}\Longleftrightarrow a = b = c = 1.$
This post has been edited 3 times. Last edited by Kunihiko_Chikaya, Jun 18, 2022, 9:46 PM
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sqing
42260 posts
#47
Y by
$$ab+bc+ca +\frac{3}{a+b+c}\geq 4\sqrt[4]{\left(\frac{ab+bc+ca}{3}\right)^3\cdot \frac{3}{a+b+c} }\geq  4\sqrt[4]{\frac{ ab+bc+ca }{3 }}\geq 4$$
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sqing
42260 posts
#48
Y by
Let $ a,b,c$ be positive real numbers satisfying $ a^2+b^2+c^2+abc=4 $ .Show that $$\frac{a}{1+2b}+\frac{b}{1+2c}+\frac{c}{1+2a}\geq1$$$$\frac{a}{1+b+c}+\frac{b}{1+c+a}+\frac{c}{1+a+b}\geq1$$
sqing wrote:
Let $ a,b,c$ be positive real numbers satisfying $ a^2+b^2+c^2+abc=4 $ .Show that $$ (b+c)(c+a)(a+b)\geq 4(a+b+c+abc-2) .$$
Solution of mudok:
Let $a+b+c=x, abc=z$. Then $ab+bc+ca=\frac{1}{2}(x^2+z-4)$

We need to prove: $x^3-12x+16\ge z(10-x)$

After using this: $ \frac{(x-2)^2}{4-x}\ge z$

it is sufficient to prove:

$(x^3-12x+16)(4-x)\ge (x-2)^2(10-x)$ $\iff (3-x)(x-2)^2(x+2)\ge 0$
This post has been edited 1 time. Last edited by sqing, Aug 14, 2022, 2:07 PM
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mihaig
7367 posts
#49
Y by
chien than wrote:
Let $ a;b;c$ be positive real numbers satisfying $ abc=1$
Prove that:
$ (a+b)(b+c)(c+a) \geq 4(a+b+c-1)$

Very beautiful
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sqing
42260 posts
#50
Y by
Let $ a,b,c>0 $ and $ \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1 $.Prove that
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{3}{a+b+c}\geq 2$$Let $ a,b,c>0 $ and $  \frac{a+b}c+1=\frac{b+c}a+\frac{c+a}b. $ Prove that
$$\frac{a+b}c\geq \frac12+\frac{\sqrt{17}}2
$$
szl6208 wrote:
Let $ a;b;c;d$ be positive real numbers satisfying $ abcd=1$ Prove that:
$$ (a+b)(b+c)(c+d)(d+a) \geq \frac{16}{3}(a+b+c+d-1)$$
https://artofproblemsolving.com/community/c6h422665p2389576:
Let $a,b,c,d>0$ and $abcd=1$. Prove that
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{16}{3(a+b+c+d)} \ge \frac{16}{3}$$
Attachments:
This post has been edited 2 times. Last edited by sqing, Mar 9, 2025, 3:38 AM
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sqing
42260 posts
#51
Y by
Let $ a,b,c>0 $ and $abc=1$. Prove that
$$ (a+2b)(b+2c)(c+2a) \geq 9(a+b+c)$$$$(a + 2b)(b + 2c)(c +2 )(2a+1) \geq 27(a + b + c ) $$
This post has been edited 1 time. Last edited by sqing, Aug 15, 2024, 7:06 AM
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