Difference between revisions of "2020 IMO Problems/Problem 2"
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| − | ==Problem== | + | == Problem == |
| − | The real numbers <math>a, b, c, d</math> are such that <math>a\ | + | 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> |
| − | Prove that | ||
| − | < | ||
| − | |||
== Solution == | == Solution == | ||
| Line 34: | Line 31: | ||
| − | ~ | + | ~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
Contents
Problem
The real numbers 
, 
, 
, 
are such that 
and 
. Prove that![\[(a + 2b + 3c + 4d) a^a b^b c^c d^d < 1.\]](http://latex.artofproblemsolving.com/f/8/b/f8b5f041651570b7cb1171443f4e990bd8cd991f.png)
Solution
Using Weighted AM-GM we get
![\[\frac{a\cdot a +b\cdot b +c\cdot c +d\cdot d}{a+b+c+d} \ge \sqrt[a+b+c+d]{a^a b^b c^c d^d}\]](http://latex.artofproblemsolving.com/e/a/8/ea8302818d5595365962a74f4a39cda18b4ee621.png)
![\[\implies a^a b^b c^c d^d \le a^2 +b^2 +c^2 +d^2\]](http://latex.artofproblemsolving.com/b/4/c/b4cf3aebe71d2bcfd733e0067e8844b7d4406415.png)
So, ![\[(a+2b+3c+4d) a^ab^bc^cd^d \le (a+2b+3c+4d)(a^2+b^2+c^2+d^2)\]](http://latex.artofproblemsolving.com/2/0/0/20089419d092c2e8e18d6021aaaaddccd188f25f.png)
Now notice that
![\[a+2b+3c+4d \text{ will be less then the following expressions (and the reason is written to the right)}\]](http://latex.artofproblemsolving.com/3/f/c/3fcf0c125107505610d157af3d2e65f48344c403.png)
![\[a+3b+3c+3d,\text{as } d\le b\]](http://latex.artofproblemsolving.com/7/f/6/7f68fcd2ecb874690f783efaf7b0000ed81024f1.png)
![\[3a+3b+3c+d, \text{as } d\le a\]](http://latex.artofproblemsolving.com/d/c/3/dc3b0da262b1c480db556924fbb655e0b511f146.png)
![\[3a+b+3c+3d, \text{as } b+d\le 2a\]](http://latex.artofproblemsolving.com/f/0/b/f0b6aa02b7aba6edd5480d1852ff5d8a8ce51ac3.png)
![\[3a+3b+c+3d, \text{as } 2c+d \le 2a+b\]](http://latex.artofproblemsolving.com/0/c/3/0c3a09434feb4074e21f4cb81f633d58b7fa24e1.png)
So, we get
![\[\begin{split}&~~~~(a+2b+3c+4d)(a^2+b^2+c^2+d^2) \\ &= a^2(a+2b+3c+4d)+b^2(a+2b+3c+4d)+c^2 (a+2b+3c+4d) +d^2 (a+2b+3c+4d)\\ &\le a^2(a+3b+3c+3d)+b^2(3a+b+3c+3d)+c^2 (3a+3b+c+3d) +d^2 (3a+3b+3c+d)\\ &<(a+b+c+d)^3 \\&=1\end{split}\]](http://latex.artofproblemsolving.com/6/5/a/65a78c74dd8d31067858ea863104d6e28603a7b1.png)
Now, for equality we must have 
In that case we get ![\[(a+2b+3c+4d) a^ab^bc^cd^d \le (a+2b+3c+4d)(a^2+b^2+c^2+d^2) =\frac{5}{8} <1\]](http://latex.artofproblemsolving.com/a/5/b/a5bb043b7080e9aa0706ced4c33f09ba55c789f4.png)
~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 | ||
