2020 IMO Problems/Problem 2

Problem

The real numbers $a$ , $b$ , $c$ , $d$ are such that $a \geq b \geq c \geq d > 0$ and $a + b + c + d = 1$ . Prove that $(a + 2b + 3c + 4d) a^a b^b c^c d^d < 1.$

Video solution

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

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

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

Video solution

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

2020 IMO (Problems) • Resources
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3
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2020 IMO Problems/Problem 2

Contents

Problem

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Solution

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