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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

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[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
UC Berkeley Integration Bee 2025 Bracket Rounds
Silver08   40
N 11 minutes ago by vanstraelen
Regular Round

Quarterfinals

Semifinals

3rd Place Match

Finals
40 replies
Silver08
May 9, 2025
vanstraelen
11 minutes ago
calculus
youochange   2
N 17 minutes ago by tom-nowy
$\int_{\alpha}^{\theta} \frac{d\theta}{\sqrt{cos\theta-cos\alpha}}$
2 replies
youochange
6 hours ago
tom-nowy
17 minutes ago
D1030 : An inequalitie
Dattier   0
an hour ago
Source: les dattes à Dattier
Let $0<a<b<c<d$ reals, and $n \in \mathbb N^*$.

Is it true that $a^n(b-a)+b^n(c-b)+c^n(d-c) \leq \dfrac {d^{n+1}}{n+1}$ ?
0 replies
Dattier
an hour ago
0 replies
Long and wacky inequality
Royal_mhyasd   0
an hour ago
Source: Me
Let $x, y, z$ be positive real numbers such that $x^2 + y^2 + z^2 = 12$. Find the minimum value of the following sum :
$$\sum_{cyc}\frac{(x^3+2y)^3}{3x^2yz - 16z - 8yz + 6x^2z}$$knowing that the denominators are positive real numbers.
0 replies
Royal_mhyasd
an hour ago
0 replies
Vietnamese national Olympiad 2007, problem 4
hien   16
N an hour ago by de-Kirschbaum
Given a regular 2007-gon. Find the minimal number $k$ such that: Among every $k$ vertexes of the polygon, there always exists 4 vertexes forming a convex quadrilateral such that 3 sides of the quadrilateral are also sides of the polygon.
16 replies
hien
Feb 8, 2007
de-Kirschbaum
an hour ago
2n^2+4n-1 and 3n+4 have common powers
bin_sherlo   4
N an hour ago by CM1910
Source: Türkiye 2025 JBMO TST P5
Find all positive integers $n$ such that a positive integer power of $2n^2+4n-1$ equals to a positive integer power of $3n+4$.
4 replies
bin_sherlo
Yesterday at 7:13 PM
CM1910
an hour ago
An interesting geometry
k.vasilev   19
N an hour ago by Ilikeminecraft
Source: All-Russian Olympiad 2019 grade 10 problem 4
Let $ABC$ be an acute-angled triangle with $AC<BC.$ A circle passes through $A$ and $B$ and crosses the segments $AC$ and $BC$ again at $A_1$ and $B_1$ respectively. The circumcircles of $A_1B_1C$ and $ABC$ meet each other at points $P$ and $C.$ The segments $AB_1$ and $A_1B$ intersect at $S.$ Let $Q$ and $R$ be the reflections of $S$ in the lines $CA$ and $CB$ respectively. Prove that the points $P,$ $Q,$ $R,$ and $C$ are concyclic.
19 replies
k.vasilev
Apr 23, 2019
Ilikeminecraft
an hour ago
sum (a+b)/(a^2+ab+b^2) <=2 if 1/a+1/b+1/c =3 for a,b,c>0
parmenides51   15
N 2 hours ago by AylyGayypow009
Source: 2020 Greek JBMO TST p2
Let $a,b,c$ be positive real numbers such that $\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}=3$. Prove that
$$\frac{a+b}{a^2+ab+b^2}+ \frac{b+c}{b^2+bc+c^2}+ \frac{c+a}{c^2+ca+a^2}\le 2$$When is the equality valid?
15 replies
parmenides51
Nov 14, 2020
AylyGayypow009
2 hours ago
Bosnia and Herzegovina JBMO TST 2016 Problem 3
gobathegreat   3
N 2 hours ago by Sh309had
Source: Bosnia and Herzegovina Junior Balkan Mathematical Olympiad TST 2016
Let $O$ be a center of circle which passes through vertices of quadrilateral $ABCD$, which has perpendicular diagonals. Prove that sum of distances of point $O$ to sides of quadrilateral $ABCD$ is equal to half of perimeter of $ABCD$.
3 replies
gobathegreat
Sep 16, 2018
Sh309had
2 hours ago
Power Of Factorials
Kassuno   180
N 2 hours ago by maromex
Source: IMO 2019 Problem 4
Find all pairs $(k,n)$ of positive integers such that \[ k!=(2^n-1)(2^n-2)(2^n-4)\cdots(2^n-2^{n-1}). \]Proposed by Gabriel Chicas Reyes, El Salvador
180 replies
1 viewing
Kassuno
Jul 17, 2019
maromex
2 hours ago
100 card with 43 having odd integers on them
falantrng   7
N 2 hours ago by Just1
Source: Azerbaijan JBMO TST 2018, D2 P4
In the beginning, there are $100$ cards on the table, and each card has a positive integer written on it. An odd number is written on exactly $43$ cards. Every minute, the following operation is performed: for all possible sets of $3$ cards on the table, the product of the numbers on these three cards is calculated, all the obtained results are summed, and this sum is written on a new card and placed on the table. A day later, it turns out that there is a card on the table, the number written on this card is divisible by $2^{2018}.$ Prove that one hour after the start of the process, there was a card on the table that the number written on that card is divisible by $2^{2018}.$
7 replies
falantrng
Aug 1, 2023
Just1
2 hours ago
GCD and LCM operations
BR1F1SZ   1
N 3 hours ago by WallyWalrus
Source: 2025 Francophone MO Juniors P4
Charlotte writes the integers $1,2,3,\ldots,2025$ on the board. Charlotte has two operations available: the GCD operation and the LCM operation.
[list]
[*]The GCD operation consists of choosing two integers $a$ and $b$ written on the board, erasing them, and writing the integer $\operatorname{gcd}(a, b)$.
[*]The LCM operation consists of choosing two integers $a$ and $b$ written on the board, erasing them, and writing the integer $\operatorname{lcm}(a, b)$.
[/list]
An integer $N$ is called a winning number if there exists a sequence of operations such that, at the end, the only integer left on the board is $N$. Find all winning integers among $\{1,2,3,\ldots,2025\}$ and, for each of them, determine the minimum number of GCD operations Charlotte must use.

Note: The number $\operatorname{gcd}(a, b)$ denotes the greatest common divisor of $a$ and $b$, while the number $\operatorname{lcm}(a, b)$ denotes the least common multiple of $a$ and $b$.
1 reply
BR1F1SZ
Saturday at 11:24 PM
WallyWalrus
3 hours ago
ISI UGB 2025 P1
SomeonecoolLovesMaths   6
N 3 hours ago by SomeonecoolLovesMaths
Source: ISI UGB 2025 P1
Suppose $f \colon \mathbb{R} \longrightarrow \mathbb{R}$ is differentiable and $| f'(x)| < \frac{1}{2}$ for all $x \in \mathbb{R}$. Show that for some $x_0 \in \mathbb{R}$, $f \left( x_0 \right) = x_0$.
6 replies
SomeonecoolLovesMaths
Yesterday at 11:30 AM
SomeonecoolLovesMaths
3 hours ago
Cute matrix equation
RobertRogo   3
N 6 hours ago by loup blanc
Source: "Traian Lalescu" student contest 2025, Section A, Problem 2
Find all matrices $A \in \mathcal{M}_n(\mathbb{Z})$ such that $$2025A^{2025}=A^{2024}+A^{2023}+\ldots+A$$Edit: Proposed by Marian Vasile (congrats!).
3 replies
1 viewing
RobertRogo
May 9, 2025
loup blanc
6 hours ago
Putnam 2003 A2
btilm305   9
N Apr 16, 2025 by KAME06
Let $a_1, a_2, \cdots , a_n$ and $b_1, b_2,\cdots, b_n$ be nonnegative real numbers. Show that \[(a_1a_2 \cdots a_n)^{1/n}+ (b_1b_2 \cdots b_n)^{1/n} \le ((a_1 + b_1)(a_2 + b_2) \cdots (a_n + b_n))^{1/n}\]
9 replies
btilm305
Jun 22, 2011
KAME06
Apr 16, 2025
Putnam 2003 A2
G H J
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btilm305
439 posts
#1 • 1 Y
Y by Adventure10
Let $a_1, a_2, \cdots , a_n$ and $b_1, b_2,\cdots, b_n$ be nonnegative real numbers. Show that \[(a_1a_2 \cdots a_n)^{1/n}+ (b_1b_2 \cdots b_n)^{1/n} \le ((a_1 + b_1)(a_2 + b_2) \cdots (a_n + b_n))^{1/n}\]
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Kent Merryfield
18574 posts
#2 • 5 Y
Y by mathman523, vsamc, GoodPerson, Adventure10, Sagnik123Biswas
More from my old document:

To start with, if any of these numbers is zero, one of the quantities on the left hand side is zero and the right hand side is clearly larger than the other side. So assume that all of the $a$s and $b$s are positive. For each $i, 1 \le i \le n,$ let $s_i = a_i + b_i.$ Let $\alpha_i = \frac{a_i}{s_i}$ and $\beta_i = \frac{b_i}{s_i} .$ Note that $\alpha_i + \beta_i = 1$ for all $i.$ Divide both sides of the inequality by $(s_1s_2\cdots s_n)^{1/n}.$ The result is that our inequality is equivalent to $(\alpha_1\alpha_2\cdots\alpha_n)^{1/n} + (\beta_1\beta_2\cdots\beta_n)^{1/n} \le 1.$ The arithmetic mean-geometric mean inequality (AM-GM inequality) now shows that $(\alpha_1\alpha_2\cdots\alpha_n)^{1/n} \le  \frac{\alpha_1 + \alpha_2 + \cdots + \alpha_n}n$ and that $(\beta_1\beta_2\cdots\beta_n)^{1/n} \le\frac{\beta_1 + \beta_2 + \cdots + \beta_n}n .$ So,
$(\alpha_1\alpha_2\cdots\alpha_n)^{1/n} + (\beta_1\beta_2\cdots\beta_n)^{1/n} \le  \frac{\alpha_1 + \beta_1 + \alpha_2 + \beta_2 + \cdots + \alpha_n + \beta_n}n$ $=  \frac{n}{n}  = 1$ and we are done.
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prasanna1712
109 posts
#3 • 2 Y
Y by Adventure10, Mango247
It seems trivial using Holder's extended inequality

see http://www.artofproblemsolving.com/Resources/Papers/MildorfInequalities.pdf
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mathbuzz
803 posts
#4 • 1 Y
Y by Adventure10
if any of $a_1,...,a_n,b_1,....b_n$ is zero , then the inequality is trivial.so , consider the case that all of them are positive.
let $a_i=x_i^n$ , $b_i=y_i^n$ .math=inline]$x_1,...,x_n ,y_1,...,y_n$[/math] are all [math=inline]$>0$[/math
then the inequality is equivalent to showing that,
$\prod{[x_i^n+y_i^n]} \ge (x_1x_2...x_n+y_1y_2....y_n)^n$
now , we use the following lemma--
Lemma-- $(p_1^n+1)......(p_n^n+1) \ge (p_1p_2....p_n+1)^n$ where all $p_i$'s are positive.
proof of the lemma--
let $p_1p_2...p_n+1=t$.
then by weighted AM-GM ,
we get $p_i^n+1=\frac{p_i^n}{t-1}.(t-1)+1 \ge t[\frac{p_i^{n(t-1)}}{(t-1)^{t-1}}]^{1/t}$
and the rest is trivial.
end of proof of lemma
now putting $p_i=\frac{x_i}{y_i}$ for all $i=1,2,..,n$ and we get the desired inequality :D :lol:
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sqing
42130 posts
#5 • 1 Y
Y by Adventure10
btilm305 wrote:
Let $a_1, a_2, \cdots , a_n$ and $b_1, b_2,\cdots, b_n$ be nonnegative real numbers. Show that \[(a_1a_2 \cdots a_n)^{1/n}+ (b_1b_2 \cdots b_n)^{1/n} \le ((a_1 + b_1)(a_2 + b_2) \cdots (a_n + b_n))^{1/n}\]
2019 CGMO P6:
Let $0\leq x_1\leq x_2\leq \cdots \leq x_n\leq 1 $ $(n\geq 2).$ Prove that $$\sqrt[n]{x_1x_2 \cdots x_n}+
\sqrt[n]{(1-x_1)(1-x_2)\cdots (1-x_n)}\leq \sqrt[n]{1-(x_1- x_n)^2}.$$
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djmathman
7938 posts
#6
Y by
An interesting fact about this problem: this is a special case of the Brunn-Minkowski Inequality.
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twbnrftw
248 posts
#7
Y by
If any of $a_i=0,$ the inequality trivially holds. Similarly for $b_i.$

Hence considering $a_i,b_i>0,$ we will employ $\text{AM-GM}.$ inequality,

Clearly $\frac{\frac{a_1}{a_1+b_1}+\frac{a_2}{a_2+b_2}+\cdots+\frac{a_n}{a_n+b_n}}{n}\le\left(\frac{a_1}{a_1+b_1}\cdot\frac{a_2}{a_2+b_2}\cdots\frac{a_n}{a_n+b_n}\right)^{1/n}$


and, $\frac{\frac{b_1}{a_1+b_1}+\frac{b_2}{a_2+b_2}+\cdots+\frac{b_n}{a_n+b_n}}{n}\le\left(\frac{b_1}{a_1+b_1}\cdot\frac{b_2}{a_2+b_2}\cdots\frac{b_n}{a_n+b_n}\right)^{1/n}.$

Adding both of them we get $1\le\left(\frac{(a_1a_2 \cdots a_n)^{1/n}+ (b_1b_2 \cdots b_n)^{1/n}}{(a_1 + b_1)(a_2 + b_2) \cdots (a_n + b_n))^{1/n}}\right).$

And hence this along with the trivial case gives us the desired inequality.$\quad\blacksquare$
This post has been edited 1 time. Last edited by twbnrftw, Jan 15, 2024, 4:07 AM
Reason: typo
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Sagnik123Biswas
420 posts
#8
Y by
It suffices to execute a set of reversible steps. First let us raise both sides to the $n^{th}$ power. We obtain the following: $\sum_{k=0}^{n}\binom{n}{k}(a_1a_2a_3 \dots a_n)^{\frac{k}{n}}(b_1b_2b_3 \dots b_n)^{\frac{n-k}{n}} \leq (a_1 + b_1)(a_2+b_2)(a_3+b_3) \dots (a_n + b_n)$.

Now let us define $S_k$ to be the part of the expansion of $(a_1 + b_1)(a_2+b_2)(a_3+b_3) \dots (a_n + b_n)$ which contains $k$ terms of the form "$a_i$" and $n-k$ terms of the form "$b_i$". It then follows that $(a_1 + b_1)(a_2+b_2)(a_3+b_3) \dots (a_n + b_n) = \sum_{k=0}^{n}S_k$. So it suffices to show that $S_k \geq \binom{n}{k}(a_1a_2a_3 \dots a_n)^{\frac{k}{n}}(b_1b_2b_3 \dots b_n)^{\frac{n-k}{n}}$.


Next, we define $P_k$ to be the product of all terms in the expansion of $(a_1 + b_1)(a_2+b_2)(a_3+b_3) \dots (a_n + b_n)$ which contain $k$ terms of the form "$a_i$" and $n-k$ terms of the form "$b_i$". By the $AM-GM$ inequality, we know that $S_k \geq \binom{n}{k}(P_k)^{\frac{1}{\binom{n}{k}}}$.

If we are able to show that $\binom{n}{k}(P_k)^{\frac{1}{\binom{n}{k}}} = \binom{n}{k}(a_1a_2a_3 \dots a_n)^{\frac{k}{n}}(b_1b_2b_3 \dots b_n)^{\frac{n-k}{n}}$, then we should be done. Observe that $(P_k)^{\frac{1}{\binom{n}{k}}} =  (a_1a_2a_3 \dots a_n) ^ { \frac{\binom{n-1}{k-1}}{\binom{n}{k}}}  (b_1b_2b_3 \dots b_n) ^ { \frac{\binom{n-1}{k}}{\binom{n}{k}}}$. The reason this holds is that each $a_i$ appears in $\binom{n-1}{k-1}$ different terms of the expansion and likewise each $b_i$ appears in $\binom{n-1}{k}$ terms of the expansion.


Simplifying $P_k$, we see that $\frac{\binom{n-1}{k-1}}{\binom{n}{k}} = \frac{(n-1)!}{(n-k)!(k-1)!} \times \frac{k!(n-k)!}{n!} =  \frac{k}{n}$. By Pascal's Identity, it must follow that $\frac{\binom{n-1}{k}}{\binom{n}{k}} = \frac{\binom{n}{k} - \binom{n-1}{k-1}}{\binom{n}{k}} = \frac{n-k}{n}$. Thus, it is indeed true that $\binom{n}{k}(P_k)^{\frac{1}{\binom{n}{k}}} = \binom{n}{k}(a_1a_2a_3 \dots a_n)^{\frac{k}{n}}(b_1b_2b_3 \dots b_n)^{\frac{n-k}{n}}$, thereby completing the proof.
This post has been edited 1 time. Last edited by Sagnik123Biswas, Jan 23, 2024, 1:34 PM
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sanyalarnab
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#9 • 1 Y
Y by mqoi_KOLA
Easy but cute problem :love:
By AM GM Inequality,
$$\frac{1}{n} \sum_{k=1}^n \frac{a_k}{a_k+b_k} \geq \left( \prod_{k=1}^n \frac{a_k}{a_k+b_k}\right)^{1/n}$$$$\frac{1}{n} \sum_{k=1}^n \frac{b_k}{a_k+b_k} \geq \left( \prod_{k=1}^n \frac{b_k}{a_k+b_k}\right)^{1/n}$$Now adding these two inequalities, we have
$$1=\frac{1}{n} \sum_{k=1}^n 1 \geq \left( \prod_{k=1}^n \frac{a_k}{a_k+b_k}\right)^{1/n}+\left( \prod_{k=1}^n \frac{b_k}{a_k+b_k}\right)^{1/n}$$This implies the statement to be proved. $\blacksquare$
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KAME06
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WLOG, if $a_1=0$, then we must prove $(b_1b_2...b_n)^{1/n} \le ((a_1 + b_1)(a_2 + b_2) \cdots (a_n + b_n))^{1/n} \Leftrightarrow b_1b_2...b_n \le (a_1 + b_1)(a_2 + b_2) \cdots (a_n + b_n)$, but $LHS$ is $RHS$'s addend, so the inequality is true
If they are positive reals:
Let $A_k$ the set of all possible products, with $n$ factors, made by $k$ distinct elements from $\{a_1, a_2, ..., a_n \}$ and $n-k$ distinct elements from $\{b_1, b_2, ..., b_n \}$ such there aren't $a, b$ with the same index. There are $\binom{n}{k}$ ways to select $k$ distinct elements from $\{a_1, a_2, ..., a_n \}$. We can't select same indeces and the product has $n$ factors, so there is $1$ way to select $n-k$ distinct elements from $\{b_1, b_2, ..., b_n \}$. All of that implies $|A_k|=\binom{n}{k}$.
Then using MA-MG: $\sum_{x \in A_k} x \ge \binom{n}{k} \sqrt[n]{\left (\prod_{i=1}^{n}a_i \right)^k \left(\prod_{i=1}^{n}b_i \right )^{n-k}}$
$\forall k=1, 2, ..., n$
Now, consider the product $(a_1+b_1)(a_2+b_2)...(a_n+b_n)$ notice that each addend is made by $k$ distinct $a$'s and $n-k$ distinct $b$'s, $\forall k=1, 2, ..., n$. Also, each element of $A_k$ is an addend of $(a_1+b_1)(a_2+b_2)...(a_n+b_n)$. That implies that $(a_1+b_1)(a_2+b_2)...(a_n+b_n)=\sum_{k=0}^n \sum_{x \in A_k}x$.
We conclude that, adding all the inequalities we obtained using MA-MG, for $k=0, 1, 2, ..., n$:
$$\sum_{k=0}^n \binom{n}{k} \sqrt[n]{\left (\prod_{i=1}^{n}a_i \right)^k \left(\prod_{i=1}^{n}b_i \right )^{n-k}} \le \sum_{k=0}^n \sum_{x \in A_k}x$$$$\Leftrightarrow \sum_{k=0}^n \binom{n}{k} \left (\sqrt[n]{\prod_{i=1}^{n}a_i} \right )^k \left (\sqrt[n]{\prod_{i=1}^{n}b_i}\right )^{n-k} \le (a_1+b_1)(a_2+b_2)...(a_n+b_n)$$$$\Leftrightarrow \left (\sqrt[n]{\prod_{i=1}^{n}a_i}+\sqrt[n]{\prod_{i=1}^{n}b_i} \right)^n \le (a_1+b_1)(a_2+b_2)...(a_n+b_n)$$$$\Leftrightarrow \left (a_1a_2...a_n \right)^{\frac{1}{n}}+\left(b_1b_2...b_n \right)^{\frac{1}{n}}\le ((a_1+b_1)(a_2+b_2)...(a_n+b_n))^{\frac{1}{n}} $$
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