Difference between revisions of "2020 IMO Problems/Problem 2"

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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>
 
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>
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~ftheftics

Revision as of 23:34, 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 ,

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


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

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

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

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

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


~ftheftics