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\frac{1}{5-2a}
Havu   1
N 2 hours ago by Havu
Let $a,b,c \ge \frac{1}{2}$ and $a^2+b^2+c^2=3$. Find minimum:
\[P=\frac{1}{5-2a}+\frac{1}{5-2b}+\frac{1}{5-2c}.\]
1 reply
Havu
Yesterday at 9:56 AM
Havu
2 hours ago
Cyclic points and concurrency [1st Lemoine circle]
shobber   10
N 4 hours ago by Ilikeminecraft
Source: China TST 2005
Let $\omega$ be the circumcircle of acute triangle $ABC$. Two tangents of $\omega$ from $B$ and $C$ intersect at $P$, $AP$ and $BC$ intersect at $D$. Point $E$, $F$ are on $AC$ and $AB$ such that $DE \parallel BA$ and $DF \parallel CA$.
(1) Prove that $F,B,C,E$ are concyclic.

(2) Denote $A_{1}$ the centre of the circle passing through $F,B,C,E$. $B_{1}$, $C_{1}$ are difined similarly. Prove that $AA_{1}$, $BB_{1}$, $CC_{1}$ are concurrent.
10 replies
shobber
Jun 27, 2006
Ilikeminecraft
4 hours ago
Hard functional equation
Jessey   4
N 4 hours ago by jasperE3
Source: Belarus 2005
Find all functions $f:N -$> $N$ that satisfy $f(m-n+f(n)) = f(m)+f(n)$, for all $m, n$$N$.
4 replies
Jessey
Mar 11, 2020
jasperE3
4 hours ago
Vertices of a convex polygon if and only if m(S) = f(n)
orl   12
N 4 hours ago by Maximilian113
Source: IMO Shortlist 2000, C3
Let $ n \geq 4$ be a fixed positive integer. Given a set $ S = \{P_1, P_2, \ldots, P_n\}$ of $ n$ points in the plane such that no three are collinear and no four concyclic, let $ a_t,$ $ 1 \leq t \leq n,$ be the number of circles $ P_iP_jP_k$ that contain $ P_t$ in their interior, and let \[m(S)=a_1+a_2+\cdots + a_n.\]Prove that there exists a positive integer $ f(n),$ depending only on $ n,$ such that the points of $ S$ are the vertices of a convex polygon if and only if $ m(S) = f(n).$
12 replies
orl
Aug 10, 2008
Maximilian113
4 hours ago
Imo Shortlist Problem
Lopes   35
N 4 hours ago by Maximilian113
Source: IMO Shortlist 2000, Problem N4
Find all triplets of positive integers $ (a,m,n)$ such that $ a^m + 1 \mid (a + 1)^n$.
35 replies
Lopes
Feb 27, 2005
Maximilian113
4 hours ago
Inspired by Humberto_Filho
sqing   0
5 hours ago
Source: Own
Let $ a,b\geq 0 $ and $a + b \leq 2$. Prove that
$$\frac{a^2+1}{(( a+ b)^2+1)^2} \geq  \frac{1}{25} $$$$\frac{(a^2+1)(b^2+1)}{((a+b)^2+1)^2} \geq  \frac{4}{25} $$$$ \frac{a^2+1}{(( a+ 2b)^2+1)^2} \geq  \frac{1}{289} $$$$ \frac{a^2+1}{((2a+ b)^2+1)^2} \geq  \frac{5}{289} $$


0 replies
sqing
5 hours ago
0 replies
Inequalities
Scientist10   2
N 5 hours ago by arqady
If $x, y, z \in \mathbb{R}$, then prove that the following inequality holds:
\[
\sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} \geq \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x
\]
2 replies
Scientist10
Yesterday at 6:36 PM
arqady
5 hours ago
$n$ with $2000$ divisors divides $2^n+1$ (IMO 2000)
Valentin Vornicu   65
N 5 hours ago by ray66
Source: IMO 2000, Problem 5, IMO Shortlist 2000, Problem N3
Does there exist a positive integer $ n$ such that $ n$ has exactly 2000 prime divisors and $ n$ divides $ 2^n + 1$?
65 replies
Valentin Vornicu
Oct 24, 2005
ray66
5 hours ago
Find the smallest of sum of elements
hlminh   0
5 hours ago
Let $S=\{1,2,...,2014\}$ and $X=\{a_1,a_2,...,a_{30}\}$ is a subset of $S$ such that if $a,b\in X,a+b\leq 2014$ then $a+b\in X.$ Find the smallest of $\dfrac{a_1+a_2+\cdots+a_{30}}{30}.$
0 replies
hlminh
5 hours ago
0 replies
Easy IMO 2023 NT
799786   133
N 5 hours ago by Maximilian113
Source: IMO 2023 P1
Determine all composite integers $n>1$ that satisfy the following property: if $d_1$, $d_2$, $\ldots$, $d_k$ are all the positive divisors of $n$ with $1 = d_1 < d_2 < \cdots < d_k = n$, then $d_i$ divides $d_{i+1} + d_{i+2}$ for every $1 \leq i \leq k - 2$.
133 replies
799786
Jul 8, 2023
Maximilian113
5 hours ago
Complicated FE
XAN4   2
N 5 hours ago by cazanova19921
Source: own
Find all solutions for the functional equation $f(xyz)+\sum_{cyc}f(\frac{yz}x)=f(x)\cdot f(y)\cdot f(z)$, in which $f$: $\mathbb R^+\rightarrow\mathbb R^+$
Note: the solution is actually quite obvious - $f(x)=x^n+\frac1{x^n}$, but the proof is important.
Note 2: it is likely that the result can be generalized into a more advanced questions, potentially involving more bash.
2 replies
XAN4
Yesterday at 11:53 AM
cazanova19921
5 hours ago
Something nice
KhuongTrang   26
N Today at 2:08 AM by KhuongTrang
Source: own
Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that

$$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le 1+2\sqrt{a+b+c+abc}.$$
26 replies
KhuongTrang
Nov 1, 2023
KhuongTrang
Today at 2:08 AM
Something nice
G H J
G H BBookmark kLocked kLocked NReply
Source: own
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KhuongTrang
729 posts
#1 • 1 Y
Y by Zuyong
Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that

$$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le 1+2\sqrt{a+b+c+abc}.$$
This post has been edited 2 times. Last edited by KhuongTrang, Nov 19, 2023, 11:59 PM
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mihaig
7343 posts
#2
Y by
Beauty. But difficult
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KhuongTrang
729 posts
#19 • 2 Y
Y by MihaiT, Zuyong
Non sense post.
This post has been edited 1 time. Last edited by KhuongTrang, Dec 23, 2023, 1:30 PM
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KhuongTrang
729 posts
#31 • 1 Y
Y by Zuyong
Something not relevant
This post has been edited 1 time. Last edited by KhuongTrang, Dec 16, 2023, 3:48 AM
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arqady
30212 posts
#32 • 1 Y
Y by teomihai
KhuongTrang wrote:
Given non-negative real numbers $a,b,c$ satisfying $ab+bc+ca=1.$ Prove that$$a\sqrt{bc+1}+b\sqrt{ca+1}+c\sqrt{ab+1}\ge 2\sqrt{a+b+c-1}.$$
Because $$\sum_{cyc}a\sqrt{bc+1}=\sqrt{\sum_{cyc}(a^2bc+a^2+2ab\sqrt{(bc+1)(ac+1)}}\geq\sqrt{\sum_{cyc}(a^2+2ab)}=a+b+c\geq2\sqrt{a+b+c-1}.$$
This post has been edited 1 time. Last edited by arqady, Nov 19, 2023, 6:25 AM
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KhuongTrang
729 posts
#34 • 1 Y
Y by Zuyong
Something not relevant
This post has been edited 2 times. Last edited by KhuongTrang, Dec 16, 2023, 3:48 AM
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sqing
41799 posts
#35
Y by
Given non-negative real numbers $a,b,c$ satisfying $ab+bc+ca=1.$ Prove that$$\sqrt{a+b+abc}+\sqrt{b+c+abc}+\sqrt{c+a+abc}\ge  2+\sqrt{2}$$
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KhuongTrang
729 posts
#43 • 1 Y
Y by Zuyong
Something not relevant
This post has been edited 1 time. Last edited by KhuongTrang, Dec 16, 2023, 3:48 AM
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KhuongTrang
729 posts
#59 • 1 Y
Y by Zuyong
Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that
$$\color{blue}{\sqrt{\frac{a+bc}{a+1}}+\sqrt{\frac{b+ca}{b+1}}+\sqrt{\frac{c+ab}{c+1}}\le 1+\sqrt{2}. }$$
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mudok
3377 posts
#60 • 1 Y
Y by arqady
arqady wrote:
KhuongTrang wrote:
Given non-negative real numbers $a,b,c$ satisfying $ab+bc+ca=1.$ Prove that$$a\sqrt{bc+1}+b\sqrt{ca+1}+c\sqrt{ab+1}\ge 2\sqrt{a+b+c-1}.$$
Because $$\sum_{cyc}a\sqrt{bc+1}=\sqrt{\sum_{cyc}(a^2bc+a^2+2ab\sqrt{(bc+1)(ac+1)}}\geq\sqrt{\sum_{cyc}(a^2+2ab)}=a+b+c\geq2\sqrt{a+b+c-1}.$$
We can directly use: $\sum a\sqrt{bc+1}\ge \sum a$ :lol:
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KhuongTrang
729 posts
#65 • 1 Y
Y by Zuyong
Problem. Given $a,b,c$ be non-negative real numbers such that $a+b+c=2.$ Prove that

$$\color{blue}{\sqrt{15a+1} +\sqrt{15b+1} +\sqrt{15c+1}\ge 3\sqrt{3}\cdot\sqrt{1+2(ab+bc+ca)}. }$$
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arqady
30212 posts
#66
Y by
KhuongTrang wrote:
Problem. Given $a,b,c$ be non-negative real numbers such that $a+b+c=2.$ Prove that

$$\color{blue}{\sqrt{15a+1} +\sqrt{15b+1} +\sqrt{15c+1}\ge 3\sqrt{3}\cdot\sqrt{1+2(ab+bc+ca)}. }$$
Holder with $(3a+1)^3$ and $uvw$.
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KhuongTrang
729 posts
#72 • 2 Y
Y by ehuseyinyigit, Zuyong
KhuongTrang wrote:
Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that
$$\color{blue}{\sqrt{\frac{a+bc}{a+1}}+\sqrt{\frac{b+ca}{b+1}}+\sqrt{\frac{c+ab}{c+1}}\le 1+\sqrt{2}. }$$

Problem. Given non-negative real numbers satisfying $ab+bc+ca=1.$ Prove that
$$\color{blue}{\sqrt{\frac{a+b}{c+1}}+\sqrt{\frac{c+b}{a+1}}+\sqrt{\frac{a+c}{b+1}}\le 2\sqrt{a+b+c}. }$$Equality holds iff $a=b=1,c=0$ or $a=b\rightarrow 0,c\rightarrow +\infty$ and any cyclic permutations.
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arqady
30212 posts
#73
Y by
KhuongTrang wrote:
Problem. Given non-negative real numbers satisfying $ab+bc+ca=1.$ Prove that
$$\color{blue}{\sqrt{\frac{a+b}{c+1}}+\sqrt{\frac{c+b}{a+1}}+\sqrt{\frac{a+c}{b+1}}\le 2\sqrt{a+b+c}. }$$Equality holds iff $a=b=1,c=0$ or $a=b\rightarrow 0,c\rightarrow +\infty$ and any cyclic permutations.
Because $$\sum_{cyc}\sqrt{\frac{a+b}{c+1}}\leq\sqrt{\sum_{cyc}(a+b)\sum_{cyc}\frac{1}{c+1}}\leq2\sqrt{a+b+c}.$$:-D
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bellahuangcat
253 posts
#74
Y by
KhuongTrang wrote:
Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that

$$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le 1+2\sqrt{a+b+c+abc}.$$

what why does that look so easy and difficult at the same time lol
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ehuseyinyigit
810 posts
#75
Y by
That's the beauty of it
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bellahuangcat
253 posts
#76
Y by
ehuseyinyigit wrote:
That's the beauty of it

yeah ig
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arqady
30212 posts
#78
Y by
sqing wrote:
Given non-negative real numbers $a,b,c$ satisfying $ab+bc+ca=1.$ Prove that$$\sqrt{a+b+abc}+\sqrt{b+c+abc}+\sqrt{c+a+abc}\ge  2+\sqrt{2}$$
The following inequality is also true.
Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc=1$. Prove that:
$$\sqrt{a+b+\frac{13}{14}abc}+\sqrt{b+c+\frac{13}{14}abc}+\sqrt{c+a+\frac{13}{14}abc}\ge  2+\sqrt{2}$$
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KhuongTrang
729 posts
#83 • 1 Y
Y by Zuyong
Problem. Given $a,b,c$ be non-negative real numbers such that $a+b+c=3.$ Prove that

$$\color{blue}{\sqrt{\frac{4}{3}(ab+bc+ca)+5}\ge \sqrt{a}+\sqrt{b}+\sqrt{c}.}$$
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kiyoras_2001
674 posts
#84
Y by
KhuongTrang wrote:
Problem. Given $a,b,c$ be non-negative real numbers such that $a+b+c=3.$ Prove that
$$\color{blue}{\sqrt{\frac{4}{3}(ab+bc+ca)+5}\ge \sqrt{a}+\sqrt{b}+\sqrt{c}.}$$
After homogenizing and squaring it becomes
\[\sum a^2+8\sum ab\ge 3\sum a\sum\sqrt{ab}.\]Changing \(a\to a^2, b\to b^2, c\to c^2\) it becomes a fourth degree inequality, so is linear in \(w^3\). Thus it remains to check only the cases \(c=0\) and \(b=c=1\) which is easy.
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KhuongTrang
729 posts
#92 • 1 Y
Y by Zuyong
Problem. Let $a,b,c$ be non-negative real variables with $ab+bc+ca>0.$ Prove that$$\color{black}{\frac{a^2+2ab}{4ab+bc+ca}+\frac{b^2+2bc}{4bc+ca+ab}+\frac{c^2+2ca}{4ca+ab+bc}\ge \frac{3}{2}. }$$Equality holds iff $(a,b,c)\sim(t,t,t)$ or $(a,b,c)\sim\left(t,0,2t\right)$ where $t>0.$
This post has been edited 1 time. Last edited by KhuongTrang, Mar 28, 2025, 1:21 AM
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jokehim
1028 posts
#93
Y by
KhuongTrang wrote:
Problem. Let $a,b,c$ be non-negative real variables with $a+b+c>0.$ Prove that$$\color{black}{\frac{a^2+2ab}{4ab+bc+ca}+\frac{b^2+2bc}{4bc+ca+ab}+\frac{c^2+2ca}{4ca+ab+bc}\ge \frac{3}{2}. }$$Equality holds iff $(a,b,c)\sim(t,t,t)$ or $(a,b,c)\sim\left(t,0,2t\right)$ where $t>0.$

Assume that $a+b+c=1$ and set $M=a^2b+b^2c+c^2a,\ \ ab+bc+ca=q,\ \ abc=r.$ The inequality becomes$$10 M^2 - 16 M q + 12 M r - 8 q^3 + 8 q^2 - 51 q r + 63 r^2 + 10 r\ge 0$$ưhere$$\Delta_M=8 (40 q^3 - 8 q^2 + 207 q r - r (297 r + 50))<0$$which ends the proofs :D
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KhuongTrang
729 posts
#94 • 1 Y
Y by Zuyong
#93 Could you please check your solution again, jokehim? I think this inequality is very hard to think of a proof in normal way.
Hope to see some ideas. Btw, it is obviously true by BW.
This post has been edited 2 times. Last edited by KhuongTrang, Mar 29, 2025, 12:05 AM
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jokehim
1028 posts
#95
Y by
KhuongTrang wrote:
#93 Could you please check your solution again, jokehim? I think this inequality is very hard to think of a proof in normal way.
Hope to see some ideas. Btw, it is obviously true by BW.
Problem. Let $a,b,c$ be positive real variables with $a+b+c+2\sqrt{abc}=1.$ Prove that$$\frac{\sqrt{a+ab+b}}{\sqrt{ab}+\sqrt{c}}+\frac{\sqrt{b+bc+c}}{\sqrt{bc}+\sqrt{a}}+\frac{\sqrt{c+ca+a}}{\sqrt{ca}+\sqrt{b}}\ge 3.$$Equality holds iff $a=b=c=\dfrac{1}{4}.$

I don't see what's wrong with my solution :|
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Nguyenhuyen_AG
3316 posts
#96
Y by
KhuongTrang wrote:
Problem. Let $a,b,c$ be non-negative real variables with $a+b+c>0.$ Prove that$$\color{black}{\frac{a^2+2ab}{4ab+bc+ca}+\frac{b^2+2bc}{4bc+ca+ab}+\frac{c^2+2ca}{4ca+ab+bc}\ge \frac{3}{2}. }$$Equality holds iff $(a,b,c)\sim(t,t,t)$ or $(a,b,c)\sim\left(t,0,2t\right)$ where $t>0.$
We have the following estimate
\[\frac{12a(a+2b)}{4ab+bc+ca} \geqslant \frac{32a^3+3(33b+56c)a^2+3(26b^2+102bc+13c^2)a-4(4b+c)(b-2c)^2}{11[ab(a+b)+bc(b+c)+ca(c+a)]+51abc}.\]
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KhuongTrang
729 posts
#109 • 2 Y
Y by arqady, Zuyong
Problem. Let $a,b,c$ be three non-negative real numbers with $ab+bc+ca=1.$ Prove that$$\frac{\sqrt{b+c}}{a+\sqrt{bc+1}}+\frac{\sqrt{c+a}}{b+\sqrt{ca+1}}+\frac{\sqrt{a+b}}{c+\sqrt{ab+1}}\ge \sqrt{2(a+b+c)}.$$When does equality hold?
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KhuongTrang
729 posts
#111
Y by
Problem. Let $a,b,c$ be three non-negative real numbers with $a+b+c=3.$ Prove that$$\color{blue}{\frac{35}{11}\ge \frac{1}{2ab+1}+\frac{1}{2bc+1}+\frac{1}{2ca+1}+\frac{4}{9}(ab+bc+ca)\ge \frac{7}{3}}.$$When does equality hold?
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