2020 IMO Problems/Problem 2

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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 $(a+2b+3c+4d)a^a b^bc^cd^d<1$


Using Weighted AM-GM we get

\[\frac{a\cdot a +b\cdot b +c\cdot c +d\cdot d}{a+b+c+d} \ge \sqrt[a+b+c+d]{a^a b^b c^c d^d}\]

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

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

Now notice that

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

So, we get \[\begin{split}&~~~~(a+2b+3c+4d)(a^2+b^2+c^2+d^2) \\ &= a^2(a+2b+3c+4d)+b^2(a+2b+3c+4d)+c^2 (a+2b+3c+4d) +d^2 (a+2b+3c+4d)\\  &\le a^2(a+3b+3c+3d)+b^2(3a+b+3c+3d)+c^2 (3a+3b+c+3d) +d^2 (3a+3b+3c+d)\\ &<(a+b+c+d)^3 \\&=1\end{split}\]

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

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


Video solution

https://youtu.be/bDHtM1wijbY [Video covers all day 1 problems]

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