Difference between revisions of "2020 IMO Problems/Problem 2"
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== Solution == | == Solution == | ||
| − | Using Weighted AM -GM we get | + | Using Weighted AM-GM we get |
<cmath>\frac{a. a +b. b +c. c +d. d}{a+b+c+d} \ge (a^a b^b c^c d^d)^{\frac{1}{a+b+c+d}}</cmath> | <cmath>\frac{a. a +b. b +c. c +d. d}{a+b+c+d} \ge (a^a b^b c^c d^d)^{\frac{1}{a+b+c+d}}</cmath> | ||
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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^c \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 | + | <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+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> |
| − | <cmath>3a+b+3c+3d , \text{as} b+d\le 2a </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> | + | <cmath>3a+3b+c+3d, \text{as } 2c+d \le 2a+b </cmath> |
| − | So, | + | So, we get |
<cmath>(a+2b+3c+4d)(a^2+b^2+c^2+d^2) </cmath> | <cmath>(a+2b+3c+4d)(a^2+b^2+c^2+d^2) </cmath> | ||
<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> | <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> | ||
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<cmath>=(a+b+c+d)^3 =1</cmath> | <cmath>=(a+b+c+d)^3 =1</cmath> | ||
| − | Now , | + | 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> | |
~ftheftics | ~ftheftics | ||
Revision as of 19:33, 27 September 2020
Problem 2. The real numbers 
are such that 
and 
.
Prove that
Solution
Using Weighted AM-GM we get
![\[\frac{a. a +b. b +c. c +d. d}{a+b+c+d} \ge (a^a b^b c^c d^d)^{\frac{1}{a+b+c+d}}\]](http://latex.artofproblemsolving.com/a/a/e/aaececd7b9f9972121408b16780174290a755ba5.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, ![\[(a+2b+3c+4d) a^ab^bc^c \le (a+2b+3c+4d)(a^2+b^2+c^2+d^2)\]](http://latex.artofproblemsolving.com/c/4/2/c42a78336c6862f35356674d16f0511afac88fa6.png)
Now notice that
![\[a+2b+3c+4d \text{ will be less then the following expression (and reason is written to the right)}\]](http://latex.artofproblemsolving.com/a/e/8/ae8a23a4c1adbbbb37f33cd352edad31c4666fcc.png)
![\[a+2b+3c+3d,\text{as } d\le b\]](http://latex.artofproblemsolving.com/9/7/7/97785361994baf0dc304fc5d47b935f21f57b41d.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
![\[(a+2b+3c+4d)(a^2+b^2+c^2+d^2)\]](http://latex.artofproblemsolving.com/7/4/5/74526744667bdcf2de4ebadf2e993173a9ac73ea.png)
![\[= a^2(a+2b+3c+4d)+b^2(a+2b+3c+4d)+c^2 (a+2b+3c+4d) +d^2 (a+2b+3c+4d)\]](http://latex.artofproblemsolving.com/3/6/a/36a0b1b46eedb471620ced0c256fad1d6541b495.png)
![\[\le a^2(a+3b+3c+3d)+b^2(3a+b+3c+3d)+c^2 (3a+3b+c+3d) +d^2 (3a+3b+3c+d)\]](http://latex.artofproblemsolving.com/c/3/b/c3b0f1c4a0d9d80a41af25c1cb71199f322da439.png)
![\[=(a+b+c+d)^3 =1\]](http://latex.artofproblemsolving.com/7/1/c/71cc9712b88b70670f69e8f1e7acc5c65d3592fe.png)
Now, for equality we must have 
In that case we get ![\[(a+2b+3c+4d) a^ab^bc^c \le (a+2b+3c+4d)(a^2+b^2+c^2+d^2) =\frac{5}{8} <1\]](http://latex.artofproblemsolving.com/9/6/5/9651d608e67240a1da98a01907545f360bdeb05b.png)
~ftheftics
