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inequalities of elements in set
toanrathay   1
N 4 hours ago by Lankou
Let \( m \) be a positive integer such that \( m \geq 4 \), and let the set
\[ A = \{a_1, a_2, a_3, \ldots, a_m\} \]consist of distinct positive integers not exceeding 2025. Suppose that for every \( a, b \in A \), with \( a \ne b \), if \( a + b \leq 2025 \), then \( a + b \in A \) as well. Prove that

\[ \frac{a_1 + a_2 + a_3 + \cdots + a_m}{m} \geq 1013. \]
1 reply
toanrathay
Today at 3:33 PM
Lankou
4 hours ago
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Complex Numbers Question
franklin2013   2
N Today at 1:59 PM by Xx_BABAI_xX
Hello everyone! This is one of my favorite complex numbers questions. Have fun!

$f(z)=z^{720}-z^{120}$. How many complex numbers $z$ are there such that $|z|=1$ and $f(z)$ is an integer.

Hint
2 replies
franklin2013
Apr 20, 2025
Xx_BABAI_xX
Today at 1:59 PM
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Inequalities
sqing   17
N Today at 1:26 PM by sqing
Let $ a,b,c> 0 $ and $ ab+bc+ca\leq 3abc . $ Prove that
$$ a+ b^2+c\leq a^2+ b^3+c^2 $$$$ a+ b^{11}+c\leq a^2+ b^{12}+c^2 $$
17 replies
sqing
Yesterday at 1:54 PM
sqing
Today at 1:26 PM
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Geometric inequality
ReticulatedPython   1
N Today at 12:43 PM by vanstraelen
Let $A$ and $B$ be points on a plane such that $AB=n$, where $n$ is a positive integer. Let $S$ be the set of all points $P$ such that $\frac{AP^2+BP^2}{(AP)(BP)}=c$, where $c$ is a real number. The path that $S$ traces is continuous, and the value of $c$ is minimized. Prove that $c$ is rational for all positive integers $n.$
1 reply
ReticulatedPython
Yesterday at 5:12 PM
vanstraelen
Today at 12:43 PM
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Binomial Sum
P162008   0
Today at 12:34 PM
Compute $\sum_{r=0}^{n} \sum_{k=0}^{r} (-1)^k (k + 1)(k + 2) \binom {n + 5}{r - k}$
0 replies
P162008
Today at 12:34 PM
0 replies
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Triple Sum
P162008   0
Today at 12:24 PM
Find the value of

$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{(-1)^m}{k.2^n + 2m + 1}$
0 replies
P162008
Today at 12:24 PM
0 replies
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Binomial Sum
P162008   0
Today at 12:03 PM
The numbers $p$ and $q$ are defined in the following manner:

$p = 99^{98} - \frac{99}{1} 98^{98} + \frac{99.98}{1.2} 97^{98} - \frac{99.98.97}{1.2.3} 96^{98} + .... + 99$

$q = 99^{100} - \frac{99}{1} 98^{100} + \frac{99.98}{1.2} 97^{100} - \frac{99.98.97}{1.2.3} 96^{100} + .... + 99$

If $p + q = k(99!)$ then find the value of $\frac{k}{10}.$
0 replies
P162008
Today at 12:03 PM
0 replies
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Polynomial Limit
P162008   0
Today at 11:55 AM
If $P_{n}(x) = \prod_{k=0}^{n} \left(x + \frac{1}{2^k}\right) = \sum_{k=0}^{n} a_{k} x^k$ then find the value of $\lim_{n \to \infty} \frac{a_{n - 2}}{a_{n - 4}}.$
0 replies
P162008
Today at 11:55 AM
0 replies
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Telescopic Sum
P162008   0
Today at 11:40 AM
Compute the value of $\Omega = \sum_{r=1}^{\infty} \frac{14 - 9r - 90r^2 - 36r^3}{7^r r(r + 1)(r + 2)(4r^2 - 1)}$
0 replies
P162008
Today at 11:40 AM
0 replies
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CHINA TST 2017 P6 DAY1
lingaguliguli   0
Today at 9:03 AM
When i search the china TST 2017 problem 6 day I i crossed out this lemme, but don't know to prove it, anyone have suggestion? tks
Given a fixed number n, and a prime p. Let f(x)=(x+a_1)(x+a_2)...(x+a_n) in which a_1,a_2,...a_n are positive intergers. Show that there exist an interger M so that 0<v_p((f(M))< n + v_p(n!)
0 replies
lingaguliguli
Today at 9:03 AM
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Combinatoric
spiderman0   1
N Today at 6:44 AM by MathBot101101
Let $ S = \{1, 2, 3, \ldots, 2024\}.$ Find the maximum positive integer $n \geq 2$ such that for every subset $T \subset S$ with n elements, there always exist two elements a, b in T such that:

$|\sqrt{a} - \sqrt{b}| < \frac{1}{2} \sqrt{a - b}$
1 reply
spiderman0
Yesterday at 7:46 AM
MathBot101101
Today at 6:44 AM
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Inequalities
sqing   10
N Apr 9, 2025 by sqing
Let $a,b$ be real numbers such that $ a^2+b^2+a^3 +b^3=4 . $ Prove that
$$a+b \leq 2$$Let $a,b$ be real numbers such that $a+b + a^2+b^2+a^3 +b^3=6 . $ Prove that
$$a+b \leq 2$$
10 replies
sqing
Apr 5, 2025
sqing
Apr 9, 2025
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Inequalities
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sqing
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sqing
#1 h  Apr 5, 2025, 1:10 PM
Y by
Let $a,b$ be real numbers such that $ a^2+b^2+a^3 +b^3=4 . $ Prove that
$$a+b \leq 2$$Let $a,b$ be real numbers such that $a+b + a^2+b^2+a^3 +b^3=6 . $ Prove that
$$a+b \leq 2$$
This post has been edited 1 time. Last edited by sqing, Apr 5, 2025, 1:19 PM
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sqing
#2 h  Apr 6, 2025, 1:41 PM
Y by
Let $ a, b, c, d\geq 0 , bc + d + a = 5, cd + a + b = 2 $ and $ da + b + c = 6. $ Prove that
$$\frac{3}{2}(\sqrt{13}-3) \leq a b + c d+d a \leq 6$$$$\frac{1}{2}(11-\sqrt{13}) \leq bc+ c d+d a \leq 6$$
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DAVROS
1661 posts
DAVROS
#3 h  Apr 6, 2025, 2:33 PM
Y by
sqing wrote:
Let $a,b$ be real numbers such that $a+b + a^2+b^2+a^3 +b^3=6 . $ Prove that $a+b \leq 2$
solution
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sqing
41795 posts
sqing
#4 h  Apr 7, 2025, 4:12 AM
Y by
Very very nice.Thank DAVROS.
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lbh_qys
549 posts
lbh_qys
#5 h  Apr 7, 2025, 4:30 AM
Y by
sqing wrote:
Let $a,b$ be real numbers such that $ a^2+b^2+a^3 +b^3=4 . $ Prove that
$$a+b \leq 2$$Let $a,b$ be real numbers such that $a+b + a^2+b^2+a^3 +b^3=6 . $ Prove that
$$a+b \leq 2$$

This inequality is readily derived from the following inequality:
For real numbers \(a\) and \(b\) satisfying \(a+b>2\) and any positive integer \(k\), it holds that \(a^k+b^k>2\).
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sqing
41795 posts
sqing
#6 h  Apr 7, 2025, 4:38 AM
Y by
Good.Thank lbh_qys.
This post has been edited 1 time. Last edited by sqing, Apr 7, 2025, 4:39 AM
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sqing
41795 posts
sqing
#7 h  Apr 7, 2025, 4:43 AM
Y by
Let $a,b$ be real numbers such that $ a^3 +b^3+ab=3 . $ Prove that
$$a+b \leq 2$$Let $a,b$ be real numbers such that $ a^2+b^2+a^3 +b^3+ab=5 . $ Prove that
$$a+b \leq 2$$Let $a,b$ be real numbers such that $a+b + a^2+b^2+a^3 +b^3+ab=7 . $ Prove that
$$a+b \leq 2$$
This post has been edited 1 time. Last edited by sqing, Apr 7, 2025, 4:46 AM
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sqing
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sqing
#8 h  Apr 8, 2025, 2:51 AM
Y by
Let $a,b$ be real numbers such that $a+ b\leq -4$. Prove that
$$ a^2 + b^2 +a^3 + b^3 + \frac{9}{2} ab \leq \frac{2197}{216}$$$$ a^2 + b^2 +a^3 + b^3 +5ab \leq \frac{343}{27}$$$$ a^2 + b^2 +a^3 + b^3 +6ab \leq \frac{512}{27}$$
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pieMax2713
4180 posts
pieMax2713
#9 h  Apr 8, 2025, 4:33 AM
Y by
sqing wrote:
Let $a,b$ be real numbers such that $a+ b\leq -4$. Prove that
$$ a^2 + b^2 +a^3 + b^3 +5ab \leq \frac{343}{27}$$
fakesolve
i think there is probably a way to make this rigorous
This post has been edited 2 times. Last edited by pieMax2713, Apr 8, 2025, 4:40 AM
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sqing
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sqing
#10 h  Apr 8, 2025, 8:35 AM
Y by
Thanks.
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sqing
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sqing
#11 h  Apr 9, 2025, 2:26 AM
Y by
Let $ a,b,c>1 $ and $ab+bc+ca+a+b+c \geq 36$. Prove that
$$ \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\leq a+b+c-8 $$Let $ a,b,c>1 $ and $ab+bc+ca \geq 27$. Prove that
$$ \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\leq a+b+c-8 $$
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