Replace maxima by minima (BxMO2021 P1)

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Replace maxima by minima (BxMO2021 P1)

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: BxMO2021 problem 1

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

42 posts

Jetze

#1 h May 3, 2021, 6:41 PM • 1 Y

Y by centslordm

(a) Prove that for all $a, b, c, d \in \mathbb{R}$ with $a + b + c + d = 0$ ,
$\max(a, b) + \max(a, c) + \max(a, d) + \max(b, c) + \max(b, d) + \max(c, d) \geqslant 0.$ (b) Find the largest non-negative integer $k$ such that it is possible to replace $k$ of the six maxima in this inequality by minima in such a way that the inequality still holds for all $a, b, c, d \in \mathbb{R}$ with $a + b + c + d = 0$ .

This post has been edited 2 times. Last edited by Jetze, May 15, 2021, 8:38 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

471 posts

booth

#2 h May 4, 2021, 7:56 AM • 2 Y

Y by centslordm, orangesyrup

WLOG $a\leq b\leq c\leq d$ .
Let $b=a+u$ , $c=a+u+v$ , $d=a+u+v+w$ , where $u,v, w$ are non-negative real numbers.
Thus $a+b+c+d=4a+3u+2v+w=0$ $\Longleftrightarrow$ $a=\frac{-3u-2v-w}{4}\text{.}$ So
$\begin{align*} \max (a,b)+\max (c,d)+\max (a,c)+\max (b,d)+\max (a,d)+\max (b,c)=b+2c+3d&=6a+6u+5v+3w\\ &=6\frac{-3u-2v-w}{4}+6u+5v+3w\\ &=\frac{3u+4v+3w}{2}\\ &\geq 0 \end{align*}$ , which is true.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1063 posts

#3 h May 4, 2021, 9:29 AM • 2 Y

Y by centslordm, StarLex1

Jetze wrote:

(a) Prove that for all $a, b, c, d \in \mathbb{R}$ with $a + b + c + d = 0$ ,
$\max(a, b) + \max(a, c) + \max(a, d) + \max(b, c) + \max(b, d) + \max(c, d) \geqslant 0.$ (b) Find the largest non-negative integer $k$ such that it is possible to replace $k$ of the
six maxima in this inequality by minima in such a way that the inequality still holds for all $a, b, c, d \in \mathbb{R}$
with $a + b + c + d = 0$ .

Here's a simpler solution for part (a) without having to bash.
Solution 01. WLOG $\max(a,b,c,d) = a$ . Note that
$4a \ge a + b + c + d = 0 \implies a \ge 0$ Hence,
$\begin{align*} \max(a,b) + \max(a,c) + \max(a,d) + \max(b,c) + \max(b,d) + \max(c,d) &= 3a + \max(b,c) + \max(b,d) + \max(c,d) \\ &\ge 3a + b + c + d \\ &= 2a \\ &\ge 0 \end{align*}$ Solution 2. Note that $\max(x,y) \ge \frac{1}{2} (x + y)$ for any $x,y \in \mathbb{R}$ . Applying this to the original inequality,
$\max(a,b) + \max(c,d) \ge 0, \max(a,c) + \max(b,d) \ge 0, \max(a,d) + \max(b,c) \ge 0$

We claim that $k = 2$ . We will prove that
$\min(a,b) + \min(c,d) + \max(a,c) + \max(a,d) + \max(b,c) + \max(b,d)\ge 0$ for any $a,b,c,d$ such that $a + b + c + d = 0$ .
To see this, note that $\max(a,b) = \frac{1}{2} (a + b + |a - b|)$ and $\min(a,b) = \frac{1}{2} (a + b - |a - b|)$ for any $a,b$ .
It suffices to prove that
$|a - c| + |a - d| + |b - c| + |b - d| \ge |a - b| + |c - d|$ which is obvious by Triangle Inequality.
To prove that $k \le 2$ , if $k = 3$ , then by PHP, some number appear twice in $\min(x,y)$ , let it be $a$ . Then let $a = -3000$ , $b = c = d = 1000$ . This gives us
$\text{Value} \le -6000 + 4(1000) = -2000 < 0$ a contradiction

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

11100 posts

#4 h Sep 15, 2021, 6:14 PM

Y by

(a). WLOG $a\ge b\ge c\ge d$ . Then:
$\text{LHS}=3a+2b+c\ge\frac32a+\frac32b+\frac32c+\frac32d=0.$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a