Art of Problem Solving
AoPS Online AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy AoPS Academy
Small live classes for advanced math
and language arts learners in grades 1-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

We have your learning goals covered with Spring and Summer courses available. Enroll today!
inequalities
B
No tags match your search
M
G
Topic
First Poster
Last Poster
No topics here!
H
J
H
Inspired by IMO 1984
sqing   2
N Mar 26, 2025 by sqing
Source: Own
Let $ a,b,c\geq 0 $ and $a+b+c=1$. Prove that
$$a^2+b^2+ ab +17abc\leq\frac{8000}{7803}$$$$a^2+b^2+ ab +\frac{163}{10}abc\leq\frac{7189057}{7173630}$$$$a^2+b^2+ ab +16.23442238abc\le1$$
2 replies
sqing
Mar 25, 2025
sqing
Mar 26, 2025
J
Inspired by IMO 1984
G H J
Reveal topic tags
inequalities
inequalities proposed
algebra
L
G H BBookmark kLocked kLocked NReply
Source: Own
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sqing
41310 posts
sqing
#1 h  Mar 25, 2025, 3:04 PM • 1 Y
Y by RainbowJessa
Let $ a,b,c\geq 0 $ and $a+b+c=1$. Prove that
$$a^2+b^2+ ab +17abc\leq\frac{8000}{7803}$$$$a^2+b^2+ ab +\frac{163}{10}abc\leq\frac{7189057}{7173630}$$$$a^2+b^2+ ab +16.23442238abc\le1$$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
awesomeming327.
1677 posts
awesomeming327.
#2 h  Mar 25, 2025, 10:50 PM • 2 Y
Y by RainbowJessa, TrendCrusher
Let $k=16.2344238$.

We have
\begin{align*} a^2+b^2+ab + kab(1-a-b) &= (a+b)^2 + ab(k(1-a-b)-1) \\ &\le (a+b)^2 + \frac{1}{4}(a+b)^2 (k-1-k(a+b))\\ &= -\frac{1}{4}k(a+b)^3 + (a+b)^2\left(\frac14 k + \frac34 \right) \end{align*}
Let $f(x)=-\frac{1}{4}kx^3 + x^2\left(\frac14 k + \frac34 \right)$. Where the domain is $[0,1]$

Since $f$ is differentiable, we simply need to find points where $f'(x)=0$, or on boundary cases, then check those points to see that $f(x)\le 1$.
\[f'(x)=-\frac{3}{4}kx^{2}+x\left(\frac{1}{2}k+\frac{3}{2}\right)\]which has solutions at $0$ and $\tfrac{2}{k}+\tfrac{2}{3}$. We have $f(0)=0$ and $f(1)=\tfrac34$ so we only need to prove that $f\left(\tfrac{2}{k}+\tfrac{2}{3}\right)\le 1$. Plugging this value back into the original expression for $f$ we get
\[(k + 3)^3\le 27 k^2\]which can be verified using something idk
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sqing
41310 posts
sqing
#3 h  Mar 26, 2025, 2:26 AM
Y by
Good.Thanks.
Z K Y
N Quick Reply
G
H
=
a