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BMN is equilateral iff rectangle ABCD is square
parmenides51   4
N an hour ago by Tsikaloudakis
Source: 2004 Romania NMO SL - Shortlist VII-VIII p8 https://artofproblemsolving.com/community/c3950157_
Consider a point $M$ on the diagonal $BD$ of a given rectangle $ABCD$, such that $\angle AMC = \angle CMD$. The point $N$ is the intersection point between $AM$ and the parallel line to $CM$ that contains $B$. Prove that the triangle $BMN$ is equilateral if and only if $ABCD$ is a square.

Valentin Vornicu
4 replies
parmenides51
Sep 16, 2024
Tsikaloudakis
an hour ago
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Loop of Logarithms
scls140511   10
N an hour ago by SomeonecoolLovesMaths
Source: 2024 China Round 1 (Gao Lian)
Round 1

1 Real number $m>1$ satisfies $\log_9 \log_8 m =2024$. Find the value of $\log_3 \log_2 m$.
10 replies
scls140511
Sep 8, 2024
SomeonecoolLovesMaths
an hour ago
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Proving a kite
Bugi   4
N an hour ago by ali123456
Source: Serbian JBTST 3, Day 2
Let $ ABCD$ be a convex quadrilateral, such that

$ \angle CBD=2\cdot\angle ADB, \angle ABD=2\cdot\angle CDB$ and $ AB=CB$.

Prove that quadrilateral $ ABCD$ is a kite.
4 replies
Bugi
May 31, 2009
ali123456
an hour ago
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Inequality
Marinchoo   6
N 2 hours ago by sqing
If $abc=1$ prove that $8(a^3+b^3+c^3) \geq 3(a^2+bc)(b^2+ac)(c^2+ab)$
6 replies
Marinchoo
Apr 28, 2020
sqing
2 hours ago
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Interesting inequality
sqing   4
N 2 hours ago by sqing
Source: Own
Let $ a,b,c\geq 2 . $ Prove that
$$(a^2-1)(b-1)(c^2-1) -\frac{9}{4}abc\geq -9$$$$(a^2-1)(b-1)(c^2-1) -\frac{11}{5}abc\geq -\frac{43}{5}$$$$(a^2-1)(b-1)(c^2-1) -2abc\geq -7$$$$(a-1)(b^2-1)(c-1) -\frac{3}{4}abc\geq -3$$$$(a-1)(b^2-1)(c-1) -\frac{3}{5}abc\geq -\frac{9}{5}$$$$(a-1)(b^2-1)(c-1) -\frac{1}{2}abc\geq -1$$
4 replies
sqing
3 hours ago
sqing
2 hours ago
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Orthocentre is collinear with two tangent points
vladimir92   42
N 2 hours ago by AshAuktober
Source: Chinese MO 1996
Let $\triangle{ABC}$ be a triangle with orthocentre $H$. The tangent lines from $A$ to the circle with diameter $BC$ touch this circle at $P$ and $Q$. Prove that $H,P$ and $Q$ are collinear.
42 replies
vladimir92
Jul 29, 2010
AshAuktober
2 hours ago
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Problem 4
den_thewhitelion   3
N 2 hours ago by DensSv
Source: Second Romanian JBMO TST 2016
We have a 4x4 board.All 1x1 squares are white.A move is changing colours of all squares of a 1x3 rectangle from black to white and from white to black.It is possible to make all the 1x1 squares black after several moves?
3 replies
den_thewhitelion
Jun 15, 2016
DensSv
2 hours ago
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Find the period
Anto0110   2
N 2 hours ago by YaoAOPS
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.
2 replies
Anto0110
Yesterday at 7:37 PM
YaoAOPS
2 hours ago
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Very interesting inequality
sqing   0
2 hours ago
Source: Own
Let $ a,b,c\geq 2 . $ Prove that
$$(a-1)(b^2-2)(c^3-3)- \frac{5}{2}abc\geq -10$$$$(a-\frac{3}{2})(b^2-2)(c^3-3)- \frac{5}{2}abc\geq -15$$$$(a-\frac{3}{2})(b^2-\frac{3}{2})(c^3-3)- \frac{25}{8}abc\geq - \frac{155}{8}$$$$(a-\frac{3}{2})(b^2-\frac{3}{2})(c^3-3)- 3abc\geq - \frac{363}{20}$$$$(a-\frac{3}{2})(b^2-\frac{3}{2})(c^3-\frac{5}{2})- \frac{55}{16}abc\geq - \frac{341}{16}$$
0 replies
sqing
2 hours ago
0 replies
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inequality
senku23   3
N 2 hours ago by SunnyEvan
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).
3 replies
senku23
5 hours ago
SunnyEvan
2 hours ago
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Inequalites
Mario16   17
N Mar 17, 2025 by sqing
If a+b+c=3 ;a,b,c>=0 prove that 1/(5+a^2)+1/(5+b^2)+1/(5+c^2)<=1/2
17 replies
Mario16
Feb 1, 2021
sqing
Mar 17, 2025
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Inequalites
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Mario16
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#1 h  Feb 1, 2021, 10:08 PM • 1 Y
Y by Mango247
If a+b+c=3 ;a,b,c>=0 prove that 1/(5+a^2)+1/(5+b^2)+1/(5+c^2)<=1/2
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IAmTheHazard
5000 posts
IAmTheHazard
#2 h  Feb 1, 2021, 10:46 PM
Y by
You literally posted this 2 hours ago.
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Mario16
100 posts
Mario16
#3 h  Feb 1, 2021, 10:50 PM
Y by
Yes but i forgot to write Something
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IAmTheHazard
5000 posts
IAmTheHazard
#4 h  Feb 1, 2021, 10:52 PM
Y by
It seems to be the exact same to me.
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Wildabandon
506 posts
Wildabandon
#5 h  Feb 2, 2021, 12:22 AM
Y by
Mario16 wrote:
Yes but i forgot to write Something

You can edit your post.

If $a,b,c\ge 0$ and $a+b+c=3$, prove that
\[\frac{1}{5+a^2} + \frac{1}{5+b^2} + \frac{1}{5+c^2}\le \frac{1}{2}\]
This post has been edited 1 time. Last edited by Wildabandon, Feb 2, 2021, 12:23 AM
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Quantum_fluctuations
1282 posts
Quantum_fluctuations
#6 h  Feb 2, 2021, 12:23 AM • 1 Y
Y by Mango247
This is where you should go.
https://artofproblemsolving.com/community/c6h2387664
This post has been edited 1 time. Last edited by Quantum_fluctuations, Feb 2, 2021, 12:23 AM
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KP9
27 posts
KP9
#7 h  Feb 2, 2021, 8:38 AM
Y by
Another solution :

Easy to prove : $\frac{1}{5+a^2} \leq \frac{1}{5}(1-\frac{1}{18}a^3 - \frac{1}{9}a)$

So we have : $\sum \frac{1}{5+a^2} \leq \frac{3}{5} - \frac{1}{90}(a^3+b^3+c^3) - \frac{1}{45}(a+b+c) = \frac{3}{5} -\frac{3}{45} - \frac{1}{90}(a^3+b^3+c^3) = \frac{8}{15} - \frac{a^3+b^3+c^3}{90}$ (1)

We also have : $(1+1+1)(1+1+1)(a^3+b^3+c^3)\geq (a+b+c)^3$

$\Rightarrow a^3 + b^3 + c^3 \geq 3$ (2)

(1) and (2) $\Rightarrow \sum \frac{1}{5+a^2} \leq \frac{8}{15} - \frac{3}{90} = \frac{1}{2}$ ( Q.E.D)
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Quantum_fluctuations
1282 posts
Quantum_fluctuations
#8 h  Feb 2, 2021, 8:43 AM
Y by
KP9 wrote:
Another solution :

Easy to prove : $\frac{1}{5+a^2} \leq \frac{1}{5}(1-\frac{1}{18}a^3 - \frac{1}{9}a)$

How did you find that?
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KP9
27 posts
KP9
#10 h  Feb 2, 2021, 9:00 AM • 1 Y
Y by Mango247
oh , iam sorry , i have a problem when i prove it
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logrange
120 posts
logrange
#11 h  Feb 2, 2021, 9:00 AM • 1 Y
Y by Mango247
I have another similar problem in which sum is cyclic in one variable only. In these kind of problems we can use tangent line, but I want to know that whether it can be used in problems which have expression ≤ constant. I have used it only in cases like expression ≥ constant. If you can do this by tangent line then please post the solution of this by tangent line also.
Prove that cyclic sum $\frac{a}{2a^2+a+1}\leq \frac{3}{4}$
Given a+b+c=3 (Sorry, I missed that earlier)
This post has been edited 3 times. Last edited by logrange, Feb 2, 2021, 5:39 PM
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logrange
120 posts
logrange
#12 h  Feb 2, 2021, 5:40 PM
Y by
Bump bump bump
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logrange
120 posts
logrange
#13 h  Feb 3, 2021, 5:07 AM
Y by
Anyone?
Note - I want a solution without n-1 EV (Calculus)
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Wildabandon
506 posts
Wildabandon
#14 h  Feb 3, 2021, 5:21 AM
Y by
I'm thinking Jensen but still using the calculus LOL
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logrange
120 posts
logrange
#15 h  Feb 3, 2021, 5:25 AM
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@below
Thanks
This post has been edited 1 time. Last edited by logrange, Feb 3, 2021, 11:13 AM
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starchan
1601 posts
starchan
#16 h  Feb 3, 2021, 5:36 AM
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I think Chebyshev's kills this one..
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sqing
41104 posts
sqing
#17 h  Mar 17, 2025, 2:31 PM
Y by
Wildabandon wrote:
If $a,b,c\ge 0$ and $a+b+c=3$, prove that
\[\frac{1}{5+a^2} + \frac{1}{5+b^2} + \frac{1}{5+c^2}\le \frac{1}{2}\]
https://artofproblemsolving.com/community/c4h2391785p19635446
https://artofproblemsolving.com/community/c6h2387664p20486022
Let $a,b,c$ be non-negative numbers such that $ab+bc+ca+abc=4.$ Prove that
$$\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}= 1$$$$\frac{1}{2}\leq \frac{1}{a^2+4}+\frac{1}{b^2+4}+\frac{1}{c^2+4}\leq \frac{3}{5}$$https://artofproblemsolving.com/community/c6h1510436p8962718
Let $ a,b,c>0 $ and $a^2+b^2+c^2+ab+bc+ca =6.$ Prove that
$$\frac{1}{a^2+5}+\frac{1}{b^2+5}+\frac{1}{c^2+5}\leq \frac{1}{2}$$( Vasile Cîrtoaje)
$$a^2b+b^2c+c^2a\leq \frac{368}{3}-\frac{176\sqrt{33}}{9}$$https://artofproblemsolving.com/community/c6h382474p2119615
This post has been edited 2 times. Last edited by sqing, Mar 17, 2025, 2:55 PM
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sqing
41104 posts
sqing
#18 h  Mar 17, 2025, 2:50 PM
Y by
Let $ a,b,c>0 $ and $a^2+b^2+c^2+ab+bc+ca =6.$ Prove that
$$ ab+bc+ca- abc\leq 2$$$$\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\geq \frac{3}{2}$$$$\frac{1}{a^2+8}+\frac{1}{b^2+8}+\frac{1}{c^2+8}\leq \frac{1}{3}$$
This post has been edited 2 times. Last edited by sqing, Mar 17, 2025, 3:30 PM
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sqing
41104 posts
sqing
#19 h  Mar 17, 2025, 3:21 PM
Y by
Let $ a,b\geq 0 $ and $a+b+a^2+ab+b^2 =5.$ Prove that
$$ \frac{1}{a^2+1}+ \frac{1}{b^2+1} \geq1$$$$ \frac{1}{a^2+2}+ \frac{1}{b^2+2} \geq \frac{2}{3}$$$$ \frac{1}{a^2+\frac{53}{20}}+ \frac{1}{b^2+\frac{53}{20}} \geq \frac{40}{73}$$$$ \frac{1}{a^2+\frac{1327}{500}}+ \frac{1}{b^2+ \frac{1327}{500}} \geq \frac{1000}{1827}$$$$ \frac{1}{a^2+3}+ \frac{1}{b^2+3} \geq \frac{185+3\sqrt{21}}{402}$$
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