Art of Problem Solving
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
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2020 IMO Problems/Problem 2"

(Created page with "Problem 2. The real numbers a, b, c, d are such that a ≥ b ≥ c ≥ d > 0 and a + b + c + d = 1. Prove that (a+2b+3c+4d)<math>a^a</math>")
 
(→‎Solution)
 
(15 intermediate revisions by 8 users not shown)
Line 1: Line 1:
−
Problem 2. The real numbers a, b, c, d are such that a ≥ b ≥ c ≥ d > 0 and a + b + c + d = 1.
+
== Problem ==
−
Prove that
+
The real numbers <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math> are such that <math>a \geq b \geq c \geq d > 0</math> and <math>a + b + c + d = 1</math>. Prove that<cmath>(a + 2b + 3c + 4d) a^a b^b c^c d^d < 1.</cmath>
−
(a+2b+3c+4d)<math>a^a</math>
+
 
 +
== Solution ==
 +
 
 +
Using Weighted AM-GM we get
 +
 
 +
<cmath>\frac{a\cdot a +b\cdot b +c\cdot c +d\cdot d}{a+b+c+d} \ge \sqrt[a+b+c+d]{a^a b^b c^c d^d}</cmath>
 +
 
 +
<cmath>\implies a^a b^b c^c d^d \le a^2 +b^2 +c^2 +d^2</cmath>
 +
 
 +
So, <cmath>(a+2b+3c+4d) a^ab^bc^cd^d \le (a+2b+3c+4d)(a^2+b^2+c^2+d^2) </cmath>
 +
 
 +
Now notice that
 +
    <cmath>a+2b+3c+4d \text{ will be less then the following expressions (and the reason is written to the right)} </cmath>
 +
    <cmath>a+3b+3c+3d,\text{as } d\le b</cmath>
 +
    <cmath>3a+3b+3c+d,              \text{as }  d\le a</cmath>
 +
    <cmath>3a+b+3c+3d, \text{as }  b+d\le 2a </cmath>
 +
  <cmath>3a+3b+c+3d, \text{as } 2c+d \le 2a+b </cmath>
 +
 
 +
 
 +
So, we get
 +
<cmath>\begin{split}&~~~~(a+2b+3c+4d)(a^2+b^2+c^2+d^2) \\
 +
&= a^2(a+2b+3c+4d)+b^2(a+2b+3c+4d)+c^2 (a+2b+3c+4d) +d^2 (a+2b+3c+4d)\\
 +
&\le a^2(a+3b+3c+3d)+b^2(3a+b+3c+3d)+c^2 (3a+3b+c+3d) +d^2 (3a+3b+3c+d)\\
 +
&<(a+b+c+d)^3 \\&=1\end{split}</cmath>
 +
 
 +
Now, for equality we must have <math>a=b=c=d=\frac{1}{4}</math>
 +
 
 +
In that case we get <cmath>(a+2b+3c+4d) a^ab^bc^cd^d \le (a+2b+3c+4d)(a^2+b^2+c^2+d^2) =\frac{5}{8} <1</cmath>
 +
 
 +
 
 +
~Shen Kislay kai
 +
 
 +
== Video solution ==
 +
 
 +
https://youtu.be/bDHtM1wijbY [Video covers all day 1 problems]
 +
 
 +
==See Also==
 +
 
 +
{{IMO box|year=2020|num-b=1|num-a=3}}
 +
 
 +
[[Category:Olympiad Algebra Problems]]

Latest revision as of 12:14, 3 September 2024

Problem

The real numbers $a$, $b$, $c$, $d$ are such that $a \geq b \geq c \geq d > 0$ and $a + b + c + d = 1$. Prove that\[(a + 2b + 3c + 4d) a^a b^b c^c d^d < 1.\]

Solution

Using Weighted AM-GM we get

\[\frac{a\cdot a +b\cdot b +c\cdot c +d\cdot d}{a+b+c+d} \ge \sqrt[a+b+c+d]{a^a b^b c^c d^d}\]

\[\implies a^a b^b c^c d^d \le a^2 +b^2 +c^2 +d^2\]

So, \[(a+2b+3c+4d) a^ab^bc^cd^d \le (a+2b+3c+4d)(a^2+b^2+c^2+d^2)\]

Now notice that 

   \[a+2b+3c+4d \text{ will be less then the following expressions (and the reason is written to the right)}\]
   \[a+3b+3c+3d,\text{as } d\le b\]
   \[3a+3b+3c+d,              \text{as }  d\le a\]
   \[3a+b+3c+3d, \text{as }  b+d\le 2a\]
 \[3a+3b+c+3d, \text{as } 2c+d \le 2a+b\]


So, we get \[\begin{split}&~~~~(a+2b+3c+4d)(a^2+b^2+c^2+d^2) \\ &= a^2(a+2b+3c+4d)+b^2(a+2b+3c+4d)+c^2 (a+2b+3c+4d) +d^2 (a+2b+3c+4d)\\ &\le a^2(a+3b+3c+3d)+b^2(3a+b+3c+3d)+c^2 (3a+3b+c+3d) +d^2 (3a+3b+3c+d)\\ &<(a+b+c+d)^3 \\&=1\end{split}\]

Now, for equality we must have $a=b=c=d=\frac{1}{4}$

In that case we get \[(a+2b+3c+4d) a^ab^bc^cd^d \le (a+2b+3c+4d)(a^2+b^2+c^2+d^2) =\frac{5}{8} <1\]


~Shen Kislay kai

Video solution

https://youtu.be/bDHtM1wijbY [Video covers all day 1 problems]

See Also

2020 IMO (Problems) • Resources
Preceded by
Problem 1 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 3
All IMO Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_IMO_Problems/Problem_2&oldid=226304"