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Inspired by IMO 1984
sqing   3
N 31 minutes ago by sqing
Source: Own
Let $ a,b,c\geq 0 $ and $a^2+b^2+ ab +24abc\geq\frac{81}{64}$. Prove that
$$a+b+\frac{9}{5}c\geq\frac{9}{8}$$$$a+b+\frac{3}{2}c\geq \frac{9}{8}\sqrt [3]{\frac{3}{2}}-\frac{3}{16}$$$$a+b+\frac{8}{5}c\geq \frac{9\sqrt [3]{25}-4}{20}$$Let $ a,b,c\geq 0 $ and $ a^2+b^2+ ab +18abc\geq\frac{343}{324} $. Prove that
$$a+b+\frac{6}{5}c\geq\frac{7\sqrt 7}{18}$$$$a+b+\frac{27}{25}c\geq\frac{35\sqrt [3]5-9}{50}$$
3 replies
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sqing
5 hours ago
sqing
31 minutes ago
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tangent circles
george_54   0
42 minutes ago
$ABC$ is a triangle with circumcenter $(\Omega)$ and $(\omega)$ is a circle tangent to $BC$ and internally to $(\Omega).$ The tangent
from $A$ to $(\omega)$ intersects $(\Omega)$ again at $D.$ If $T, P$ are the contact points of $(\omega)$ with $BC, AD$ respectively, prove that $CT=AC\cdot PD+DC\cdot PA.$
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george_54
42 minutes ago
0 replies
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2025 Caucasus MO Seniors P6
BR1F1SZ   1
N 44 minutes ago by pco
Source: Caucasus MO
It is known that from segments of lengths $a$, $b$ and $c$, a triangle can be formed. Can it happen that from segments of lengths$$\sqrt{a^2 + bc}, \quad \sqrt{b^2 + ca}, \quad \sqrt{c^2 + ab}$$an obtuse triangle can be formed?
1 reply
BR1F1SZ
Today at 12:48 AM
pco
44 minutes ago
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Integers with determinant \pm 1
anantmudgal09   32
N an hour ago by anudeep
Source: INMO 2021 Problem 1
Suppose $r \ge 2$ is an integer, and let $m_1, n_1, m_2, n_2, \dots, m_r, n_r$ be $2r$ integers such that $$\left|m_in_j-m_jn_i\right|=1$$for any two integers $i$ and $j$ satisfying $1 \le i<j \le r$. Determine the maximum possible value of $r$.

Proposed by B Sury
32 replies
anantmudgal09
Mar 7, 2021
anudeep
an hour ago
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Long polynomial factorization
wassupevery1   0
an hour ago
Source: 2025 Vietnam IMO TST - Problem 6
For each prime $p$ of the form $4k+3$ with $k \in \mathbb{Z}^+$, consider the polynomial $$Q(x)=px^{2p} - x^{2p-1} + p^2x^{\frac{3p+1}{2}}+2(p^2+1)x^p + p^2 x^{\frac{p-1}{2}} -x + p.$$Determine all ordered pairs of polynomials $f, g$ with integer coefficients such that $Q(x)=f(x)g(x)$.
0 replies
+1 w
wassupevery1
an hour ago
0 replies
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Good set of cells
wassupevery1   0
an hour ago
Source: 2025 Vietnam IMO TST - Problem 5
There is an $n \times n$ grid which has rows and columns numbered from $1$ to $n$; the cell at row $i$ and column $j$ is denoted as the cell at $(i, j)$. A subset $A$ of the cells is called good if for any two cells at $(x_1, y), (x_2, y)$, the cells $(u, v)$ satisfying $x_1 < u \leq x_2, v<y$ or $x_1 \leq u < x_2, v>y$ are not in $A$. Determine the minimal number of good sets such that they are pairwise disjoint and every cell of the board belongs to exactly one good set.
0 replies
wassupevery1
an hour ago
0 replies
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Find the value
sqing   2
N an hour ago by sqing
Source: Hunan changsha 2025
Let $ a,b,c $ be real numbers such that $ abc\neq 0,2a-b+c= 0 $ and $ a-2b-c=0. $ Find the value of $\frac{a^2+b^2+c^2}{ab+bc+ca}.$
Let $ a,b,c $ be real numbers such that $ abc\neq 0,a+2b+3c= 0 $ and $ 2a+3b+4c=0. $ Find the value of $\frac{ab+bc+ca}{a^2+b^2+c^2}.$
2 replies
sqing
3 hours ago
sqing
an hour ago
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Binomial Non-divisibility
wassupevery1   0
an hour ago
Source: 2025 Vietnam IMO TST - Problem 4
Find all positive integers $k$ for which there are infinitely many positive integers $n$ such that $\binom{(2025+k)n}{2025n}$ is not divisible by $kn+1$.
0 replies
wassupevery1
an hour ago
0 replies
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Inspired by IMO 1984
sqing   2
N an hour ago by sqing
Source: Own
Let $ a,b,c\geq 0 $ and $a+b+c=1$. Prove that
$$a^2+b^2+ ab +24abc\leq\frac{81}{64}$$Equality holds when $a=b=\frac{3}{8},c=\frac{1}{4}.$
$$a^2+b^2+ ab +18abc\leq\frac{343}{324}$$Equality holds when $a=b=\frac{7}{18},c=\frac{2}{9}.$
2 replies
sqing
6 hours ago
sqing
an hour ago
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2025 Caucasus MO Juniors P1
BR1F1SZ   1
N an hour ago by pco
Source: Caucasus MO
Anya and Vanya’s houses are located on the straight road. The distance between their houses is divided by a shop and a school into three equal parts. If Anya and Vanya leave their houses at the same time and walk towards each other, they will meet near the shop. If Anya rides a scooter, then her speed will increase by $150\,\text{m/min}$, and they will meet near the school. Find Vanya’s speed of walking.
1 reply
BR1F1SZ
Today at 12:54 AM
pco
an hour ago
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Interesting inequality
sqing   7
N Mar 22, 2025 by sqing
Source: Own
Let $ a,b> 0$ and $ a+b=1 . $ Prove that
$$ \frac{1}{a}+\frac{1}{b}\geq \frac{2k}{1+k^2 a^2b^2}$$Where $ 4\leq k\in N^+.$
$$ \frac{1}{a}+\frac{1}{b}\geq \frac{4+\frac{k}{4}}{1+ ka^2b^2}$$Where $16\geq k>0 .$
7 replies
sqing
Mar 22, 2025
sqing
Mar 22, 2025
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Interesting inequality
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Reveal topic tags
inequalities proposed
inequalities
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G H BBookmark kLocked kLocked NReply
Source: Own
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sqing
41249 posts
sqing
#1 h  Mar 22, 2025, 3:45 AM
Y by
Let $ a,b> 0$ and $ a+b=1 . $ Prove that
$$ \frac{1}{a}+\frac{1}{b}\geq \frac{2k}{1+k^2 a^2b^2}$$Where $ 4\leq k\in N^+.$
$$ \frac{1}{a}+\frac{1}{b}\geq \frac{4+\frac{k}{4}}{1+ ka^2b^2}$$Where $16\geq k>0 .$
This post has been edited 3 times. Last edited by sqing, Mar 22, 2025, 1:52 PM
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ionbursuc
949 posts
ionbursuc
#2 h  Mar 22, 2025, 3:59 AM
Y by
sqing wrote:
Let $ a,b> 0$ and $ a+b=1 . $ Prove that
$$ \frac{1}{a}+\frac{1}{b}\geq \frac{2k}{1+k^2 a^2b^2}$$Where $ 5\leq k\in N^+.$

$\frac{1}{a}+\frac{1}{b}-\frac{2k}{1+{{k}^{2}}{{a}^{2}}{{b}^{2}}}=\frac{a+b}{ab}-\frac{2k}{1+{{k}^{2}}{{a}^{2}}{{b}^{2}}}=\frac{1+{{k}^{2}}{{a}^{2}}{{b}^{2}}-2kab}{ab\left( 1+{{k}^{2}}{{a}^{2}}{{b}^{2}} \right)}=\frac{{{\left( kab-1 \right)}^{2}}}{ab\left( 1+{{k}^{2}}{{a}^{2}}{{b}^{2}} \right)}\ge 0,\forall k\in \mathbb{R}$
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sqing
41249 posts
sqing
#3 h  Mar 22, 2025, 1:14 PM
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Nice.Thanks.
This post has been edited 1 time. Last edited by sqing, Mar 22, 2025, 1:21 PM
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sqing
41249 posts
sqing
#4 h  Mar 22, 2025, 1:19 PM
Y by
Let $ a,b> 0$ and $ a+b=1 . $ Prove that
$$ \frac{1}{a}+\frac{1}{b}\geq \frac{4+\frac{k}{4}}{1+ ka^2b^2}$$Where $16\geq k>0 .$
$$ \frac{1}{a}+\frac{1}{b}\geq \frac{\frac{17}{4}}{1+ a^2b^2}$$$$ \frac{1}{a}+\frac{1}{b}\geq \frac{5}{1+4 a^2b^2}$$$$ \frac{1}{a}+\frac{1}{b}\geq \frac{\frac{25}{4}}{1+9a^2b^2}$$
This post has been edited 3 times. Last edited by sqing, Mar 22, 2025, 1:53 PM
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SunnyEvan
39 posts
SunnyEvan
#5 h  Mar 22, 2025, 1:29 PM
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sqing wrote:
Let $ a,b> 0$ and $ a+b=1 . $ Prove that
$$ \frac{1}{a}+\frac{1}{b}\geq \frac{\frac{17}{4}}{1+ a^2b^2}$$$$ \frac{1}{a}+\frac{1}{b}\geq \frac{5}{1+4 a^2b^2}$$$$ \frac{1}{a}+\frac{1}{b}\geq \frac{\frac{25}{4}}{1+9a^2b^2}$$

$$a+b=1$$$$ \frac{1}{a}+\frac{1}{b}\geq \frac{\frac{17}{4}}{1+ a^2b^2}$$<===>$$ a^2b^2- \frac{17}{4}ab+1 \geq 0 $$<===>$$ ab \leq \frac{1}{4}$$
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sqing
41249 posts
sqing
#6 h  Mar 22, 2025, 1:48 PM
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Nice.Thanks.
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SunnyEvan
39 posts
SunnyEvan
#8 h  Mar 22, 2025, 1:54 PM
Y by
sqing wrote:
Let $ a,b> 0$ and $ a+b=1 . $ Prove that
$$ \frac{1}{a}+\frac{1}{b}\geq \frac{4+\frac{k}{4}}{1+ ka^2b^2}$$Where $16\geq k>0 .$
$$ \frac{1}{a}+\frac{1}{b}\geq \frac{\frac{17}{4}}{1+ a^2b^2}$$$$ \frac{1}{a}+\frac{1}{b}\geq \frac{5}{1+4 a^2b^2}$$$$ \frac{1}{a}+\frac{1}{b}\geq \frac{\frac{25}{4}}{1+9a^2b^2}$$

$$ \frac{1}{a}+\frac{1}{b}\geq \frac{4+\frac{k}{4}}{1+ ka^2b^2}$$$$ (4+ \frac{k}{4})ab \leq 1+ka^2b^2$$<===>$$ (kab-4)(4ab-1) \geq 0$$<===>$$ ab \leq \frac{1}{4}$$
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sqing
41249 posts
sqing
#9 h  Mar 22, 2025, 2:46 PM
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Nice.Thanks.
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