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Difference between revisions of "2020 IMO Problems/Problem 2"
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Problem 2. The real numbers a, b, c, d are such that a≥b≥c≥d>0 and <math>a+b+c+d=1</math>.
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Problem 2. The real numbers <math>a, b, c, d</math> are such that <math>a\ge b \ge c\ge d > 0</math> and <math>a+b+c+d=1</math>.
 
Prove that
 
Prove that
 
<math>(a+2b+3c+4d)a^a b^bc^cd^d<1</math>
 
<math>(a+2b+3c+4d)a^a b^bc^cd^d<1</math>
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 +
 +
== Solution ==
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 +
Using Weighted AM -GM we get,
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<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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<cmath>\implies a^a b^b c^c d^d \le a^2 +b^2 +c^2 +d^2</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>
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 +
Now notice that ,
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\[
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    a+2b+3c+4d \le
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\begin{cases}
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    a+3b+3c+3d,& \text{as }  d\le b\\
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    3a+3b+3c+d,              &\text{as}  d\le a \\
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    3a+b+3c+3d ,& \text{as}
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b+d\le 2a \\
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  3a +3b +c +3d ,& \text{as}
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2c+d \le 2a+b
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\end{cases}
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\]
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So, We get ,
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<cmath>(a+2b+3c+4d)(a^2+b^2+c^2+d^2) </cmath>
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<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>\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>
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<cmath>=(a+b+c+d)^3 =1</cmath>
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Now , For equality we must have <math>a=b=c=d=\frac{1}{4}</math>
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 +
On 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>

Revision as of 23:20, 26 September 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,

\[\frac{a. a +b. b +c. c +d. d}{a+b+c+d} \ge (a^a b^b c^c d^d)^{\frac{1}{a+b+c+d}}\]

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

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

Now notice that ,

\[ 

   a+2b+3c+4d \le

\begin{cases} 

   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


\end{cases} \]

So, We get , \[(a+2b+3c+4d)(a^2+b^2+c^2+d^2)\] \[= a^2(a+2b+3c+4d)+b^2(a+2b+3c+4d)+c^2 (a+2b+3c+4d) +d^2 (a+2b+3c+4d)\]

\[\le a^2(a+3b+3c+3d)+b^2(3a+b+3c+3d)+c^2 (3a+3b+c+3d) +d^2 (3a+3b+3c+d)\]

\[=(a+b+c+d)^3 =1\]

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

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

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