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"

m (→‎Solution)
m (→‎Solution)
Line 12: Line 12:
 
<cmath>\implies a^a b^b c^c d^d \le a^2 +b^2 +c^2 +d^2</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^c \le (a+2b+3c+4d)(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
 
Now notice that
−
     <cmath>a+2b+3c+4d \text{ will be less then the following expression (and reason is written to the right)} </cmath>
+
     <cmath>a+2b+3c+4d \text{ will be less then the following expressions (and the reason is written to the right)} </cmath>
 
     <cmath>a+2b+3c+3d,\text{as } d\le b</cmath>
 
     <cmath>a+2b+3c+3d,\text{as } d\le b</cmath>
 
     <cmath>3a+3b+3c+d,              \text{as }  d\le a</cmath>
 
     <cmath>3a+3b+3c+d,              \text{as }  d\le a</cmath>
Line 23: Line 23:
  
 
So, we get
 
So, we get
−
<cmath>(a+2b+3c+4d)(a^2+b^2+c^2+d^2) </cmath>
+
<cmath>\begin{split}&~~~~(a+2b+3c+4d)(a^2+b^2+c^2+d^2) \\
−
<cmath>= a^2(a+2b+3c+4d)+b^2(a+2b+3c+4d)+c^2 (a+2b+3c+4d) +d^2 (a+2b+3c+4d) </cmath>
+
&= 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)\\
−
<cmath>\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)</cmath>
+
&<(a+b+c+d)^3 \\&=1\end{split}</cmath>
−
 
 
−
<cmath><(a+b+c+d)^3 =1</cmath>
 
  
 
Now, for equality we must have <math>a=b=c=d=\frac{1}{4}</math>
 
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^c \le (a+2b+3c+4d)(a^2+b^2+c^2+d^2) =\frac{5}{8} <1</cmath>
+
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>
  
  
 
~ftheftics
 
~ftheftics

Revision as of 08:36, 2 October 2020

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


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 

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


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


~ftheftics

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_IMO_Problems/Problem_2&oldid=134484"