# Difference between revisions of "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

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

$$\implies a^a b^b c^c d^d \le a^2 +b^2 +c^2 +d^2$$

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

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


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

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

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

~ftheftics