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Given 
such that 
. Prove that
such that 
. Prove that
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Symmetric inequalities under two constraints
ChrP 4
N
by ChrP
Let 
such that 
. Prove that 
and
What about the lower bound in the first case and the upper bound in the second?
such that 
. Prove that 
and
What about the lower bound in the first case and the upper bound in the second?
4 replies
Terms of a same AP
adityaguharoy 1
N
by Mathzeus1024
Given 
are positive integers pairwise distinct and 
is also a positive integer 
.
Determine under which conditions can![$\sqrt[n]{p},\sqrt[n]{q},\sqrt[n]{r}$](//latex.artofproblemsolving.com/3/1/3/3133b60c921168020d149a9d28ef69b756a704d9.png)
form terms of a same arithmetic progression.
are positive integers pairwise distinct and 
is also a positive integer 
.Determine under which conditions can
![$\sqrt[n]{p},\sqrt[n]{q},\sqrt[n]{r}$](http://latex.artofproblemsolving.com/3/1/3/3133b60c921168020d149a9d28ef69b756a704d9.png)
form terms of a same arithmetic progression.1 reply
Inspired by old results
sqing 8
N
by sqing
Source: Own
Let 
and 
. Prove that
Let 
be reals such that 
. Prove that
and 
. Prove that
Let 
be reals such that 
. Prove that
8 replies
Transform the sequence
steven_zhang123 1
N
by vgtcross
Given a sequence of 
real numbers 
, we can select a real number 
and transform the sequence into 
. This transformation can be performed multiple times, with each chosen real number potentially being different
(i) Prove that it is possible to transform the sequence into all zeros after a finite number of such transformations.
(ii) To ensure that the above result can be achieved for any given initial sequence, what is the minimum number of transformations required?
real numbers 
, we can select a real number 
and transform the sequence into 
. This transformation can be performed multiple times, with each chosen real number potentially being different(i) Prove that it is possible to transform the sequence into all zeros after a finite number of such transformations.
(ii) To ensure that the above result can be achieved for any given initial sequence, what is the minimum number of transformations required?
1 reply
NEPAL TST DAY 2 PROBLEM 2
Tony_stark0094 5
N
by ThatApollo777
Kritesh manages traffic on a 
grid consisting of 2025 unit squares. Within each unit square is a car, facing either up, down, left, or right. If the square in front of a car in the direction it is facing is empty, it can choose to move forward. Each car wishes to exit the grid.
Kritesh realizes that it may not always be possible for all the cars to leave the grid. Therefore, before the process begins, he will remove
cars from the grid in such a way that it becomes possible for all the remaining cars to eventually exit the grid.
What is the minimum value of that guarantees that Kritesh's job is possible?
grid consisting of 2025 unit squares. Within each unit square is a car, facing either up, down, left, or right. If the square in front of a car in the direction it is facing is empty, it can choose to move forward. Each car wishes to exit the grid.Kritesh realizes that it may not always be possible for all the cars to leave the grid. Therefore, before the process begins, he will remove
cars from the grid in such a way that it becomes possible for all the remaining cars to eventually exit the grid.What is the minimum value of
that guarantees that Kritesh's job is possible?
5 replies
Product of distinct integers in arithmetic progression -- ever a perfect power ?
adityaguharoy 1
N
by Mathzeus1024
Source: Well known (the gen. is more difficult, but may be not this one -- so this is here)
Let 
be four positive integers in arithmetic progression (that is, 
) and with 
. Can the product 
ever be a number of the form 
for some 
and some 
, with 
?
be four positive integers in arithmetic progression (that is, 
) and with 
. Can the product 
ever be a number of the form 
for some 
and some 
, with 
?
1 reply
For positive integers \( a, b, c \), find all possible positive integer values o
Jackson0423 2
N
by ATM_
For positive integers 
, find all possible positive integer values of
![\[
\frac{a}{b} + \frac{b}{c} + \frac{c}{a}.
\]](//latex.artofproblemsolving.com/3/2/b/32b41635c159921e11245369e66185ec69b0d785.png)
, find all possible positive integer values of![\[
\frac{a}{b} + \frac{b}{c} + \frac{c}{a}.
\]](http://latex.artofproblemsolving.com/3/2/b/32b41635c159921e11245369e66185ec69b0d785.png)
2 replies
Isosceles Triangle Geo
oVlad 2
N
by SomeonesPenguin
Source: Romania Junior TST 2025 Day 1 P2
Consider the isosceles triangle 
with 
and the circle 
of radius 
centered at 
Let 
be the midpoint of 
The line 
intersects a second time at 
Let 
be a point on such that 
Let 
be the intersection of 
and Prove that 
with 
and the circle 
of radius 
centered at 
Let 
be the midpoint of 
The line 
intersects a second time at 
Let 
be a point on such that 
Let 
be the intersection of 
and Prove that 
2 replies
IMO ShortList 1998, number theory problem 5
orl 63
N
by ATM_
Source: IMO ShortList 1998, number theory problem 5
Determine all positive integers for which there exists an integer 
such that 
is a divisor of 
.
for which there exists an integer 
such that 
is a divisor of 
.
63 replies
IMO Shortlist 2013, Number Theory #1
lyukhson 150
N
by MuradSafarli
Source: IMO Shortlist 2013, Number Theory #1
Let 
be the set of positive integers. Find all functions 
such that
![\[ m^2 + f(n) \mid mf(m) +n \]](//latex.artofproblemsolving.com/2/f/4/2f409d1de993f1af8fd839bb8e9f87a57e1b8608.png)
for all positive integers and .
be the set of positive integers. Find all functions 
such that![\[ m^2 + f(n) \mid mf(m) +n \]](http://latex.artofproblemsolving.com/2/f/4/2f409d1de993f1af8fd839bb8e9f87a57e1b8608.png)
for all positive integers
and .
150 replies
An inequality problem
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Y by Arithmetic_fighter
Clearly 
yields a stronger inequality, so let 
, 
, 
. We need to show that
![\[\sum_{\rm cyc}\frac{ab}{c^2+a^2+1}=\sum_{\rm cyc}\frac{3ab}{4c^2+4a^2+b^2}\le1,\]](//latex.artofproblemsolving.com/9/c/2/9c26ff3a0d26422b2e6b8737508ce7c6ceb898ff.png)
or
![\[\sum_{\rm cyc}\left(\frac34-\frac{3ab}{4c^2+4a^2+b^2}\right)=\sum_{\rm cyc}\frac{12c^2+3(2a-b)^2}{4\left(4c^2+4a^2+b^2\right)}\ge3\times\frac34-1=\frac54.\]](//latex.artofproblemsolving.com/9/d/f/9df0e5e43eb4aa5bc7ed6840c91f1b5407f30cdc.png)
By C-S, we have
![\[\sum_{\rm cyc}\frac{12c^2+3(2a-b)^2}{4c^2+4a^2+b^2}\ge\frac{\left\{\sum\limits_{\rm cyc}\left[12c^2+3(2a-b)^2\right]\right\}^2}{\sum\limits_{\rm cyc}\left[12c^2+3(2a-b)^2\right]\left(4c^2+4a^2+b^2\right)}.\]](//latex.artofproblemsolving.com/9/7/d/97d8d081d238446616e436860ae189d00e82adfd.png)
Therefore, it suffices to show that
![\[\left\{\sum\limits_{\rm cyc}\left[12c^2+3(2a-b)^2\right]\right\}^2\ge5\sum\limits_{\rm cyc}\left[12c^2+3(2a-b)^2\right]\left(4c^2+4a^2+b^2\right).\]](//latex.artofproblemsolving.com/7/1/8/71827e7b9f7dac09b44e21ddbd0ce942c317f3ff.png)
This is a fourth degree polynomial inequality, so we have
yields a stronger inequality, so let 
, 
, 
. We need to show that![\[\sum_{\rm cyc}\frac{ab}{c^2+a^2+1}=\sum_{\rm cyc}\frac{3ab}{4c^2+4a^2+b^2}\le1,\]](http://latex.artofproblemsolving.com/9/c/2/9c26ff3a0d26422b2e6b8737508ce7c6ceb898ff.png)
or![\[\sum_{\rm cyc}\left(\frac34-\frac{3ab}{4c^2+4a^2+b^2}\right)=\sum_{\rm cyc}\frac{12c^2+3(2a-b)^2}{4\left(4c^2+4a^2+b^2\right)}\ge3\times\frac34-1=\frac54.\]](http://latex.artofproblemsolving.com/9/d/f/9df0e5e43eb4aa5bc7ed6840c91f1b5407f30cdc.png)
By C-S, we have![\[\sum_{\rm cyc}\frac{12c^2+3(2a-b)^2}{4c^2+4a^2+b^2}\ge\frac{\left\{\sum\limits_{\rm cyc}\left[12c^2+3(2a-b)^2\right]\right\}^2}{\sum\limits_{\rm cyc}\left[12c^2+3(2a-b)^2\right]\left(4c^2+4a^2+b^2\right)}.\]](http://latex.artofproblemsolving.com/9/7/d/97d8d081d238446616e436860ae189d00e82adfd.png)
Therefore, it suffices to show that![\[\left\{\sum\limits_{\rm cyc}\left[12c^2+3(2a-b)^2\right]\right\}^2\ge5\sum\limits_{\rm cyc}\left[12c^2+3(2a-b)^2\right]\left(4c^2+4a^2+b^2\right).\]](http://latex.artofproblemsolving.com/7/1/8/71827e7b9f7dac09b44e21ddbd0ce942c317f3ff.png)
This is a fourth degree polynomial inequality, so we have
![\begin{align*}
&m \sum_{\rm cyc} a^{4}+n \sum_{\rm cyc} b^{2} c^{2}+p \sum_{\rm cyc} a^{3} b+q \sum_{\rm cyc} a b^{3}-(m+n+p+q) \sum_{\rm cyc} a^2bc \\
={}&\frac{1}{18 m} \sum_{\rm cyc}\left[3 m\left(a^{2}-b^{2}\right)+(p+2 q) c a+(p-q) a b-(2 p+q) b c\right]^{2}\\
&+\frac{3 m(m+n)-\left(p^{2}+p q+q^{2}\right)}{18 m\left(p^{2}+p q+q^{2}\right)} \sum_{\rm cyc}[(p+2 q) c a+(p-q) a b-(2 p+q) b c]^{2}.
\end{align*}](http://latex.artofproblemsolving.com/7/e/f/7ef2fe85183f92f06e1decea6e259a27ec2da3bd.png)
This post has been edited 1 time. Last edited by Quantum-Phantom, Apr 3, 2025, 6:52 AM
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#3
Apr 3, 2025, 5:45 AM
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Quantum-Phantom wrote:
This is a fourth-degree polynomial inequality, so we have
Could you please clarify how you obtained this equation? It seems complicated.
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