Difference between revisions of "2020 IMO Problems/Problem 2"
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− | Problem | + | == Problem == |
− | + | 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> | |
− | < | + | |
+ | == 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> | ||
+ | |||
+ | |||
+ | ~ftheftics | ||
+ | |||
+ | == 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 21:18, 15 July 2024
Contents
Problem
The real numbers ,
,
,
are such that
and
. Prove that
Solution
Using Weighted AM-GM we get
So,
Now notice that
So, we get
Now, for equality we must have
In that case we get
~ftheftics
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 |