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Cyclic ine
m4thbl3nd3r   5
N 6 minutes ago by sqing
Let $a,b,c>0$ such that $a+b+c=3$. Prove that $$a^3b+b^3c+c^3a+9abc\le 12$$
5 replies
m4thbl3nd3r
Yesterday at 3:17 PM
sqing
6 minutes ago
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Interesting inequality
sqing   1
N 29 minutes ago by sqing
Source: Own
Let $ a,b,c\geq 0,(ab+c)(ac+b)\neq 0 $ and $ a+b+c=3 . $ Prove that
$$ \frac{1}{ab+kc}+\frac{1}{ac+kb} \geq\frac{4}{3k} $$Where $ k\geq 3. $
$$ \frac{1}{ab+2c}+\frac{1}{ac+2b} \geq\frac{16}{25} $$$$ \frac{1}{ab+3c}+\frac{1}{ac+3b} \geq\frac{4}{9} $$$$ \frac{1}{ab+4c}+\frac{1}{ac+4b} \geq\frac{1}{3} $$

1 reply
sqing
32 minutes ago
sqing
29 minutes ago
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Problem 2
delegat   145
N 31 minutes ago by Marcus_Zhang
Source: 0
Let $n\ge 3$ be an integer, and let $a_2,a_3,\ldots ,a_n$ be positive real numbers such that $a_{2}a_{3}\cdots a_{n}=1$. Prove that
\[(1 + a_2)^2 (1 + a_3)^3 \dotsm (1 + a_n)^n > n^n.\]

Proposed by Angelo Di Pasquale, Australia
145 replies
delegat
Jul 10, 2012
Marcus_Zhang
31 minutes ago
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CMJ 1284 (Crazy Concyclic Circumcenter Circus)
kgator   0
32 minutes ago
Source: College Mathematics Journal Volume 55 (2024), Issue 4: https://doi.org/10.1080/07468342.2024.2373015
1284. Proposed by Tran Quang Hung, High School for Gifted Students, Vietnam National University, Hanoi, Vietnam. Let quadrilateral $ABCD$ not be a trapezoid such that there is a circle centered at $I$ that is tangent to the four sides $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, and $\overline{DA}$. Let $X$, $Y$, $Z$, and $W$ be the circumcenters of the triangles $IAB$, $IBC$, $ICD$, and $IDA$, respectively. Prove that there is a circle containing the circumcenters of the triangles $XAB$, $YBC$, $ZCD$, and $WDA$.
0 replies
kgator
32 minutes ago
0 replies
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euler-totient function
Laan   3
N 34 minutes ago by top1vien
Proof that there are infinitely many positive integers $n$ such that
$\varphi(n)<\varphi(n+1)<\varphi(n+2)$
3 replies
Laan
Friday at 7:13 AM
top1vien
34 minutes ago
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Loop of Logarithms
scls140511   12
N 44 minutes ago by yyhloveu1314
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 $\log_3 \log_2 m$.
12 replies
scls140511
Sep 8, 2024
yyhloveu1314
44 minutes ago
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A cyclic inequality
JK1603JK   1
N an hour ago by jokehim
Source: unknown
Let a,b,c be real numbers. Prove that a^6+b^6+c^6\ge 2(a+b+c)(ab+bc+ca)(a-b)(b-c)(c-a).
1 reply
JK1603JK
3 hours ago
jokehim
an hour ago
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bank accounts
cloventeen   1
N an hour ago by jkim0656
edgar has three bank accounts, each with an integer amount of dollars in it. He is only allowed to transfer money from one account to another if, by doing so, the latter ends up with double the money it had previously. Prove that edgar can always transfer all of his money into two accounts. Will he always be able to transfer all of his money into a single account?
1 reply
cloventeen
2 hours ago
jkim0656
an hour ago
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Divisibility
RenheMiResembleRice   0
an hour ago
Source: Byer
Prove that for all n ∈ ℕ, 133|($11^{\left(n+2\right)}+12^{\left(2n+1\right)}$).
0 replies
RenheMiResembleRice
an hour ago
0 replies
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2 var inquality
sqing   3
N an hour ago by sqing
Source: Own
Let $ a,b>0 $ and $ 3a+4b=a^3b^2. $ Prove that
$$2a+b+\dfrac{2}{a}+\dfrac{3}{b}\geq \frac{11}{\sqrt2}$$$$a+\dfrac{2}{a}+\dfrac{3}{b}\geq 4\sqrt[4]{\frac23}$$$$\dfrac{2}{a}+\dfrac{3}{b}\geq 2\sqrt[4]3$$$$3a+\dfrac{2}{a}+\dfrac{3}{b}\geq \sqrt[4]{354+66\sqrt{33}}$$
3 replies
sqing
Mar 4, 2025
sqing
an hour ago
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Tangent.
steven_zhang123   1
N an hour ago by ehuseyinyigit
Source: China TST 2001 Quiz 6 P1
In \( \triangle ABC \) with \( AB > BC \), a tangent to the circumcircle of \( \triangle ABC \) at point \( B \) intersects the extension of \( AC \) at point \( D \). \( E \) is the midpoint of \( BD \), and \( AE \) intersects the circumcircle of \( \triangle ABC \) at \( F \). Prove that \( \angle CBF = \angle BDF \).
1 reply
steven_zhang123
2 hours ago
ehuseyinyigit
an hour ago
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Inequality and function
srnjbr   5
N Yesterday at 8:50 AM by pco
Find all f:R--R such that for all x,y, yf(x)+f(y)>=f(xy)
5 replies
srnjbr
Friday at 4:26 PM
pco
Yesterday at 8:50 AM
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Inequality and function
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inequalities
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srnjbr
56 posts
srnjbr
#1 h  Friday at 4:26 PM
Y by
Find all f:R--R such that for all x,y, yf(x)+f(y)>=f(xy)
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ektorasmiliotis
86 posts
ektorasmiliotis
#2 h  Friday at 5:15 PM
Y by
solution: f(x)=0
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pco
23460 posts
pco
#3 h  Friday at 5:31 PM
Y by
ektorasmiliotis wrote:
solution: f(x)=0
Plus a lot lot of others
You should post your proof
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pco
23460 posts
pco
#4 h  Friday at 5:32 PM
Y by
srnjbr wrote:
Find all f:R--R such that for all x,y, yf(x)+f(y)>=f(xy)
Let $P(x,y)$ be the assertion $yf(x)+f(y)\ge f(xy)$

Let $x,y\ne 0$ :
Adding $P(\frac xy,y)$ and $P(\frac xy,-y)$, we get $f(y)+f(-y)\ge f(x)+f(-x)$
And so (swapping $x,y$) : $f(x)+f(-x)=c$ for some constant $c$ and $\forall x\ne 0$

$P(x,-y)$ $\implies$ $-yf(x)+c-f(y)\ge c-f(xy)$ and so $f(xy)\ge yf(x)+f(y)$
And so, comparing with $P(x,y)$ $f(xy)=yf(x)+f(y)$ $\forall x,y\ne 0$

Swapping there $x,y$ and subtracting : $(y-1)f(x)=(x-1)f(y)$ $\forall x,y\ne 0$
And so $f(x)=a(x-1)$ for some constant $a$ and $\forall x\ne 0$

Let then $x\ne 0$ : $P(0,x)$ $\implies$ $xf(0)+a(x-1)\ge f(0)$ $\implies$ $(x-1)(f(0)+a)\ge 0$ and so $f(0)=-a$

And so $\boxed{f(x)=a(x-1)\quad\forall x}$ which indeed fits, whatever is $a\in\mathbb R$
This post has been edited 1 time. Last edited by pco, Yesterday at 8:50 AM
Reason: Typo
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srnjbr
56 posts
srnjbr
#5 h  Friday at 6:32 PM • 2 Y
Y by aidan0626, pco
f(0)=-a.
Thank you very much for answering
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pco
23460 posts
pco
#6 h  Yesterday at 8:50 AM
Y by
srnjbr wrote:
f(0)=-a.
Indeed :), thanks
Edited
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