2020 IMO Problems/Problem 2

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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2020 IMO Problems/Problem 2

Solution