Happy Thanksgiving! Please note that there are no classes November 25th-December 1st.

G
Topic
First Poster
Last Poster
x - 1/y = y - 1/z = z - 1/x (German MO)
aether_1729   2
N 15 minutes ago by rchokler
Find all real $x, y, z$ such that
\[x - \frac{1}{y} = y - \frac{1}{z} = z - \frac{1}{x}.\]
Source: Problem 520943 from 52nd German MO, final round / day 1, grade 9.
2 replies
aether_1729
Yesterday at 12:45 PM
rchokler
15 minutes ago
Inequalities
sqing   5
N 2 hours ago by sqing
Let $ a,b\geq1 $ and $ a+b+ab=8 .$ Prove that
$$\frac{3\sqrt{5}}{2}\geq \frac{\sqrt{a^2-1}}{b^2}+\frac{\sqrt{b^2-1}}{a^2}\geq \frac{\sqrt{3}}{2} $$
5 replies
sqing
Yesterday at 7:36 AM
sqing
2 hours ago
No more topics!
Inequalities
sqing   55
N Nov 21, 2024 by sqing
Let $a,b>0$ and $ab^2(a+b)=9.$ Prove that
$$2a+5b\geq 2\sqrt[4]{27(3+8\sqrt{6})} $$$$2a+9b\geq 6\sqrt[4]{48\sqrt{2}-39} $$
55 replies
sqing
Oct 31, 2024
sqing
Nov 21, 2024
Inequalities
G H J
G H BBookmark kLocked kLocked NReply
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sqing
39045 posts
#1
Y by
Let $a,b>0$ and $ab^2(a+b)=9.$ Prove that
$$2a+5b\geq 2\sqrt[4]{27(3+8\sqrt{6})} $$$$2a+9b\geq 6\sqrt[4]{48\sqrt{2}-39} $$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sqing
39045 posts
#2
Y by
Let $a,b>0$ and $ a+2b\leq 8.$ Prove that
$$ab^2(2a+b)\leq \frac{471+133\sqrt{57}}{9} $$Let $a,b>0$ and $a+3b\leq 4.$ Prove that
$$ab^2(a+b)\leq \frac{39+55\sqrt{33} }{144} $$
This post has been edited 1 time. Last edited by sqing, Nov 1, 2024, 3:30 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
DAVROS
1475 posts
#3 • 1 Y
Y by soryn
sqing wrote:
Let $a,b>0$ and $ab^2(a+b)=9.$ Prove that $2a+5b\geq 2\sqrt[4]{27(3+8\sqrt{6})} $
solution
This post has been edited 1 time. Last edited by DAVROS, Nov 1, 2024, 9:45 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sqing
39045 posts
#4
Y by
Very very nice.Thank DAVROS.
Let $a,b >0 .$ Prove that
$$ \frac{a^2}{b+1}+ \frac{b^2}{a+1}\geq \frac{a^2+b^2}{\sqrt{ab}+1}$$Let $a,b,c >0 .$ Prove that $$ \frac{a^3}{b^3+1}+ \frac{b^3}{c^3+1}+ \frac{c^3}{a^3+1}\geq \frac{3abc}{abc+1}$$$$ \frac{a}{(a+b)^2+c^2}+ \frac{b}{(b+c)^2+a^2}+ \frac{c}{(c+a)^2+b^2}\leq \frac{9}{5(a+b+c)}$$
This post has been edited 4 times. Last edited by sqing, Nov 17, 2024, 1:13 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sqing
39045 posts
#5
Y by
Let $ a,b>0 $ and $ a+b=2. $ Prove that
$$\left(\frac{311}{50}ab +1 \right)\left(a+\frac{1}{b} +\frac{b}{a}\right)  \geq \frac{1083}{50} $$$$\left(\frac{1247}{200}ab +1 \right)\left(a+\frac{1}{b} +\frac{b}{a}\right)  \geq \frac{4341}{200} $$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
DAVROS
1475 posts
#6
Y by
sqing wrote:
Let $a,b>0$ and $ a+2b\leq 8.$ Prove that $ab^2(2a+b)\leq \frac{471+133\sqrt{57}}{9} $
solution
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sqing
39045 posts
#7
Y by
Very very nice.Thank DAVROS.
Let $ a,b,c\geq 0 .$ Prove that$$ \frac{a}{b+c}+ \frac{b }{c+a}+ \frac{4c}{a+b+c}\geq 2$$$$ \sqrt{\frac{a}{b+c}}+  \sqrt{\frac{b }{c+a}}+  2\sqrt{\frac{c}{a+b+c}}\geq 2$$Let $ a,b>0 .$ Prove that$$ \frac{a^3}{b(a+1)^2}+ \frac{b^3}{a(b+1)^2}\geq \frac{ab}{ab+1}$$
Attachments:
This post has been edited 2 times. Last edited by sqing, Nov 17, 2024, 12:43 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sqing
39045 posts
#8
Y by
Let $ a,b>0 $ and $ (a^{12}+1)(b^{12}+1)\le 4 .$ Prove that$$ a+b+ ab   \leq 3$$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mangowo
3 posts
#9
Y by
I'm not very good at inequalities so idk if my solution works but here it is anyways:

Click to reveal hidden text
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sqing
39045 posts
#10
Y by
Thanks.
Let $ a,b,c\geq 0 .$ Prove that$$ \frac{a}{(b+c)^2}+ \frac{b }{(c+a)^2}+ \frac{4}{a+b}\geq  \frac{2(a+b+c)}{ab+bc+ca}$$Let $ a,b,c>0 $ and $ a+b+ab\geq 3.$ Prove that$$ \frac{a}{b}+ \frac{1}{a}+b\geq \frac{9}{2ab+1}$$Let $ a,b,c>0 $ and $ ab+bc+ca+abc\geq 4.$ Prove that$$ \frac{a}{b}+ \frac{c}{a}+ \frac{1}{c}+b\geq \frac{16}{3abc+1}$$
This post has been edited 1 time. Last edited by sqing, Nov 20, 2024, 3:01 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ytChen
1052 posts
#11
Y by
sqing wrote:
Let $ a,b>0 $ and $ (a^{12}+1)(b^{12}+1)\le 4 .$ Prove that$$ a+b+ ab   \leq 3$$
Solution. From the assumption and AM-GM inequality, it follows that
\begin{align*}36\ge&32+  (a^{12}+1)(b^{12}+1)\\
=& 11+a^{12}+11+b^{12}+11+(ab)^{12}\\
\ge&12a+12b+12ab,
\end{align*}thereby $a+b+ab\le3$, and the maximum value $3$ can be attained if $a=b=1$. $\blacksquare$
This post has been edited 2 times. Last edited by ytChen, Nov 2, 2024, 2:56 AM
Reason: Typo
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sqing
39045 posts
#12
Y by
Very very nice.Thank ytChen.
Let $ a,b,c\geq 0 .$ Prove that$$ \frac{a}{b+c}+ \frac{b }{c+a}+ \frac{4(c+\sqrt{ab})}{a+b}\geq 4$$$$ \frac{a}{b+c}+ \frac{b }{c+a}+ \frac{9c+\sqrt{ab}}{a+b}\geq 6$$$$ \frac{a}{b+c}+  \frac{b+\sqrt{ca}}{c+a}+ \frac{c+\sqrt{ab}}{a+b}\geq 2$$$$ \frac{a+\sqrt{bc}}{b+c}+ \frac{b+\sqrt{ca} }{c+a}+ \frac{ c+\sqrt{ab}}{a+b} \geq \frac52$$
This post has been edited 2 times. Last edited by sqing, Nov 21, 2024, 2:50 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sqing
39045 posts
#13
Y by
Let $ a,b,c\geq 0 $ and $ a+b+c=1. $ Prove that
$$ 2\leq \frac{(a^2 +1)(b^2 +1)(c^2 +1)}{a^2 +b^2 +c^2 } \leq  \frac{1000}{243}  $$$$ \frac{1000}{1701} \leq\frac{(a^2 +1)(b^2 +1)(c^2 +1)}{a^2 +b^2 +c^2 +2}\leq  \frac{2}{3}  $$
This post has been edited 1 time. Last edited by sqing, Nov 2, 2024, 5:02 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
DAVROS
1475 posts
#14
Y by
sqing wrote:
Let $ a,b,c\geq 0 $ and $ a+b+c=1. $ Prove that $ 2\leq \frac{(a^2 +1)(b^2 +1)(c^2 +1)}{a^2 +b^2 +c^2 } \leq  \frac{1000}{243}  $
solution
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sqing
39045 posts
#15
Y by
Very very nice.Thank DAVROS.
Let $ a,b> 0 $ and $ a+b=2. $ Prove that
$$ (a +\sqrt{a^2 +3b}) (b +\sqrt{b^2 +3a})\geq 9$$$$\frac{a}{b+\sqrt{a^2 +3b}}+\frac{b}{a+\sqrt{b^2 +3a}}\geq \frac{2}{3}$$$$\frac{1}{a+\sqrt{b^2 +3a}}+\frac{1}{b+\sqrt{a^2 +3b}}\geq \frac{2}{3}$$https://artofproblemsolving.com/community/c6h2338297p18827338
This post has been edited 1 time. Last edited by sqing, Nov 24, 2024, 1:20 AM
Z K Y
G
H
=
a