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A cyclic inequality
KhuongTrang   4
N 24 minutes ago by paixiao
Source: own-CRUX
IMAGE
https://cms.math.ca/.../uploads/2025/04/Wholeissue_51_4.pdf
4 replies
KhuongTrang
Apr 21, 2025
paixiao
24 minutes ago
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Perfect polynomials
Phorphyrion   5
N 2 hours ago by Davdav1232
Source: 2023 Israel TST Test 5 P3
Given a polynomial $P$ and a positive integer $k$, we denote the $k$-fold composition of $P$ by $P^{\circ k}$. A polynomial $P$ with real coefficients is called perfect if for each integer $n$ there is a positive integer $k$ so that $P^{\circ k}(n)$ is an integer. Is it true that for each perfect polynomial $P$, there exists a positive $m$ so that for each integer $n$ there is $0<k\leq m$ for which $P^{\circ k}(n)$ is an integer?
5 replies
Phorphyrion
Mar 23, 2023
Davdav1232
2 hours ago
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Finding all integers with a divisibility condition
Tintarn   14
N 2 hours ago by Assassino9931
Source: Germany 2020, Problem 4
Determine all positive integers $n$ for which there exists a positive integer $d$ with the property that $n$ is divisible by $d$ and $n^2+d^2$ is divisible by $d^2n+1$.
14 replies
Tintarn
Jun 22, 2020
Assassino9931
2 hours ago
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Geometry Handout is finally done!
SimplisticFormulas   2
N 3 hours ago by parmenides51
If there’s any typo or problem you think will be a nice addition, do send here!
handout, geometry
2 replies
1 viewing
SimplisticFormulas
Yesterday at 4:58 PM
parmenides51
3 hours ago
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functional equation interesting
skellyrah   4
N 3 hours ago by skellyrah
find all functions IR->IR such that $$xf(x+yf(xy)) + f(f(y)) = f(xf(y))^2 + (x+1)f(x)$$
4 replies
skellyrah
4 hours ago
skellyrah
3 hours ago
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IMO ShortList 2002, number theory problem 2
orl   57
N 3 hours ago by Maximilian113
Source: IMO ShortList 2002, number theory problem 2
Let $n\geq2$ be a positive integer, with divisors $1=d_1<d_2<\,\ldots<d_k=n$. Prove that $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$ is always less than $n^2$, and determine when it is a divisor of $n^2$.
57 replies
orl
Sep 28, 2004
Maximilian113
3 hours ago
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"Mistakes were made" -Luke Rbotaille
a1267ab   10
N 3 hours ago by Martin.s
Source: USA TST 2025
Let $a_1, a_2, \dots$ and $b_1, b_2, \dots$ be sequences of real numbers for which $a_1 > b_1$ and
\begin{align*} a_{n+1} &= a_n^2 - 2b_n\\ b_{n+1} &= b_n^2 - 2a_n \end{align*}for all positive integers $n$. Prove that $a_1, a_2, \dots$ is eventually increasing (that is, there exists a positive integer $N$ for which $a_k < a_{k+1}$ for all $k > N$).

Holden Mui
10 replies
a1267ab
Dec 14, 2024
Martin.s
3 hours ago
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Problem 4 (second day)
darij grinberg   92
N 3 hours ago by cubres
Source: IMO 2004 Athens
Let $n \geq 3$ be an integer. Let $t_1$, $t_2$, ..., $t_n$ be positive real numbers such that \[n^2 + 1 > \left( t_1 + t_2 + \cdots + t_n \right) \left( \frac{1}{t_1} + \frac{1}{t_2} + \cdots + \frac{1}{t_n} \right).\] Show that $t_i$, $t_j$, $t_k$ are side lengths of a triangle for all $i$, $j$, $k$ with $1 \leq i < j < k \leq n$.
92 replies
darij grinberg
Jul 13, 2004
cubres
3 hours ago
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Perpendicularity with Incircle Chord
tastymath75025   31
N 4 hours ago by cj13609517288
Source: 2019 ELMO Shortlist G3
Let $\triangle ABC$ be an acute triangle with incenter $I$ and circumcenter $O$. The incircle touches sides $BC,CA,$ and $AB$ at $D,E,$ and $F$ respectively, and $A'$ is the reflection of $A$ over $O$. The circumcircles of $ABC$ and $A'EF$ meet at $G$, and the circumcircles of $AMG$ and $A'EF$ meet at a point $H\neq G$, where $M$ is the midpoint of $EF$. Prove that if $GH$ and $EF$ meet at $T$, then $DT\perp EF$.

Proposed by Ankit Bisain
31 replies
tastymath75025
Jun 27, 2019
cj13609517288
4 hours ago
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\frac{1}{5-2a}
Havu   2
N 4 hours ago by arqady
Let $a\ge b\ge c \ge \frac{1}{2}$ and $a^2+b^2+c^2=3$. Find minimum:
\[P=\frac{1}{5-2a}+\frac{1}{5-2b}+\frac{1}{5-2c}.\]
2 replies
Havu
Wednesday at 9:56 AM
arqady
4 hours ago
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\frac{1}{9}+\frac{1}{\sqrt{3}}\geq a^2+\sqrt{a+ b^2} \geq \frac{1}{4}
sqing   2
N 5 hours ago by zaidova
Source: Own
Let $a,b\geq 0 $ and $3a+4b =1 .$ Prove that
$$\frac{2}{3}\geq a +\sqrt{a^2+ 4b^2}\geq \frac{6}{13}$$$$\frac{1}{9}+\frac{1}{\sqrt{3}}\geq a^2+\sqrt{a+ b^2} \geq \frac{1}{4}$$$$2\geq a+\sqrt{a^2+16b} \geq \frac{2}{3}\geq a+\sqrt{a^2+16b^3} \geq \frac{2(725-8\sqrt{259})}{729}$$
2 replies
sqing
Oct 3, 2023
zaidova
5 hours ago
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Interesting inequalities
sqing   6
N Apr 16, 2025 by sqing
Source: Own
Let $ a,b,c\geq 0 $ and $ ab+bc+ca+abc=4$ . Prove that
$$k(a+b+c) -ab-bc\geq 4\sqrt{k(k+1)}-(k+4)$$Where $ k\geq \frac{16}{9}. $
$$ \frac{16}{9}(a+b+c) -ab-bc\geq \frac{28}{9}$$
6 replies
sqing
Apr 16, 2025
sqing
Apr 16, 2025
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Interesting inequalities
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inequalities
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Source: Own
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sqing
41809 posts
sqing
#1 h  Apr 16, 2025, 3:36 AM
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Let $ a,b,c\geq 0 $ and $ ab+bc+ca+abc=4$ . Prove that
$$k(a+b+c) -ab-bc\geq 4\sqrt{k(k+1)}-(k+4)$$Where $ k\geq \frac{16}{9}. $
$$ \frac{16}{9}(a+b+c) -ab-bc\geq \frac{28}{9}$$
This post has been edited 1 time. Last edited by sqing, Apr 16, 2025, 6:13 AM
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41809 posts
sqing
#2 h  Apr 16, 2025, 3:37 AM
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Let $ a,b,c\geq 0 $ and $ ab+bc+ca+abc =4$ . Prove that
$$k(a+b+c) -ab-bc\geq 4(k-1) $$Where $ \frac{16}{9}\geq k\geq 0. $
This post has been edited 1 time. Last edited by sqing, Apr 16, 2025, 6:16 AM
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#3 h  Apr 16, 2025, 4:02 AM
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Let $ a,b,c\geq 0 $ and $ ab+bc+ca+abc =4$ . Prove that
$$k(a+b+c) +a^2-bc\geq 4(k-1) $$Where $ 2.6647417 \geq k\geq 0. $
$$ k(a+b+c) +a^2-ab - bc\geq 4(k-1) $$Where $ 1.968711\geq k\geq 0. $
This post has been edited 2 times. Last edited by sqing, Apr 16, 2025, 6:18 AM
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lbh_qys
549 posts
lbh_qys
#4 h  Apr 16, 2025, 5:47 AM
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sqing wrote:
Let $ a,b,c\geq 0 $ and $ a^2+b^2+c^2+abc=4$ . Prove that
$$k(a+b+c) -ab-bc\geq 4\sqrt{k(k+1)}-(k+4)$$Where $ k\geq \frac{16}{9}. $
$$ \frac{16}{9}(a+b+c) -ab-bc\geq \frac{28}{9}$$

When $a=b=\sqrt{2}, c=0$, then $\frac{16}{9}(a+b+c) -ab-bc= \frac{32\sqrt{2}}{9} - 2 < \frac{28}{9}$.
This post has been edited 1 time. Last edited by lbh_qys, Apr 16, 2025, 5:47 AM
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sqing
41809 posts
sqing
#5 h  Apr 16, 2025, 6:14 AM
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Sorry .Thank lbh_qys.
Let $ a,b,c\geq 0 $ and $ ab+bc+ca+abc=4$ . Prove that
$$k(a+b+c) -ab-bc\geq 4\sqrt{k(k+1)}-(k+4)$$Where $ k\geq \frac{16}{9}. $
$$ \frac{16}{9}(a+b+c) -ab-bc\geq \frac{28}{9}$$
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lbh_qys
549 posts
lbh_qys
#6 h  Apr 16, 2025, 10:13 AM
Y by
sqing wrote:
Let $ a,b,c\geq 0 $ and $ ab+bc+ca+abc=4$ . Prove that
$$k(a+b+c) -ab-bc\geq 4\sqrt{k(k+1)}-(k+4)$$Where $ k\geq \frac{16}{9}. $
$$ \frac{16}{9}(a+b+c) -ab-bc\geq \frac{28}{9}$$
sqing wrote:
Let $ a,b,c\geq 0 $ and $ ab+bc+ca+abc =4$ . Prove that
$$k(a+b+c) -ab-bc\geq 4(k-1) $$Where $ \frac{16}{9}\geq k\geq 0. $


If \(b > k\), since \(ab+bc \leq ab+bc+ca+abc = 4\), we have \( a+c \le \frac{4}{b}\), therefore

\[ k(a+b+c) - ab - bc + (k+4) = kb + (a+c)(k-b) + k + 4 \ge kb + \frac{4}{b}(k-b) + k + 4 \]\[ = kb + \frac{4k}{b} + k \ge 2\sqrt{kb \cdot \frac{4k}{b}} + k = 5k \]
If \(b \le k\), since
\[ 0 = b(a+c) + (b+1)ac - 4 \le b(a+c) + \frac{(b+1)(a+c)^2}{4} - 4 = (a+c+4)\bigl(a+c + ab + bc - 4\bigr), \]
we have \( a+c \geq \frac{4}{1+b} \), thus
\[ k(a+b+c) - ab - bc + (k+4) = k(b+1) + (a+c)(k-b) + 4 \ge k(b+1) + \frac{4}{b+1}(k-b) + 4 \]\[ = k(b+1) + \frac{4(k+1)}{b+1} \ge 2\sqrt{ k(b+1) \cdot \frac{4(k+1)}{b+1}} = 4 \sqrt{k(k+1)} \]
Hence we have

\[ k(a+b+c) -ab-bc \geq \min\left\{5k, 4\sqrt{k(k+1)} \right\} -(k+4) \]
This post has been edited 8 times. Last edited by lbh_qys, Apr 16, 2025, 10:57 AM
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#7 h  Apr 16, 2025, 12:29 PM
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Very very nice.Thank lbh_qys.
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