IMO 2020 Problem 2

Proposed by: -
Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 1m15s

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

Exploration

The expression splits into two components: a linear form in $a,b,c,d$ and a multiplicative entropy-type term $a^a b^b c^c d^d$. The constraint $a \ge b \ge c \ge d > 0$ with $a+b+c+d=1$ forces an ordered probability vector.

The multiplicative term suggests the function $x \mapsto x \log x$, which is convex on $(0,\infty)$, hence $\sum x_i \log x_i$ behaves in a controlled way under averaging constraints. The linear term suggests rearrangement inequality, since coefficients $1,2,3,4$ are increasing while $(a,b,c,d)$ is decreasing.

A natural strategy is to decouple the problem into a bound for the linear expression and a bound for the entropy expression, then combine them. The most delicate point is that neither factor is uniformly small; control must come from their interaction.

A key expected mechanism is that extremal behavior occurs when the vector is as unbalanced as allowed by ordering, or when it is fully symmetric. The symmetric point $a=b=c=d=\tfrac14$ gives a strictly small value of the product, indicating the inequality is far from tight there.

The main difficulty lies in preventing configurations where the linear factor becomes large while the entropy factor remains close to $1$.

Problem Understanding

This is a Type C problem, requiring proof that a certain expression is always strictly less than $1$ under ordering and normalization constraints.

We are given positive real numbers $a \ge b \ge c \ge d$ summing to $1$, and we must show

$$(a+2b+3c+4d), a^a b^b c^c d^d < 1.$$

The structure is a product of a weighted sum and an entropy-like multiplicative term. The linear factor favors placing mass on larger coefficients, while the product term penalizes dispersion in a nonlinear way.

The expected mechanism is a combination of rearrangement to control the linear factor and convexity of $x \log x$ to control the multiplicative factor. The difficulty is that both factors can increase in opposing regimes, so a naive independent bound is insufficient.

The inequality is expected to be strict due to positivity constraints preventing degeneracy.

Proof Architecture

Define $S = a+2b+3c+4d$ and $P = a^a b^b c^c d^d$.

First, a rearrangement-based lemma bounds $S$ in terms of the oppositely ordered pairing of coefficients, producing an upper bound depending linearly on $a,b,c,d$.

Second, a convexity lemma for $x \log x$ gives a universal upper bound on $P$ via exponential representation, namely $P = \exp(\sum x \log x)$ and convexity of $x \log x$.

Third, a combined inequality lemma shows that the product $SP$ is maximized under a structured extremal configuration consistent with ordering constraints.

The hardest step is the interaction between the rearranged linear term and the entropy bound, since neither is independently tight in the same regime.

Solution

Lemma 1

For real numbers $a \ge b \ge c \ge d$ and coefficients $1 \le 2 \le 3 \le 4$, the rearrangement inequality gives

$$a+2b+3c+4d \le 4a+3b+2c+d.$$

The rearrangement inequality states that for two sequences ordered oppositely, the sum of pairwise products is minimized. Applying it to $(a,b,c,d)$ and $(1,2,3,4)$ yields the inequality above. ∎

This step establishes that the worst case for the linear factor occurs when large variables are paired with large coefficients.

We now rewrite the right-hand side using $a+b+c+d=1$:

$$4a+3b+2c+d = (a+b+c+d) + 3a+2b+c = 1 + 3a+2b+c.$$

Hence

$$a+2b+3c+4d \le 1 + 3a+2b+c.$$

Lemma 2

For every $x>0$, $\log x \le x-1$, with equality only at $x=1$.

This follows from convexity of $\log x$ or by considering $f(x)=x-1-\log x$ and verifying $f'(x)=1-\tfrac1x$ and $f''(x)=\tfrac1{x^2}>0$, hence $f$ attains its minimum $0$ at $x=1$. ∎

Multiplying by $x>0$ gives

$$x \log x \le x(x-1).$$

Summing over $a,b,c,d$ yields

$$\sum x \log x \le a(a-1)+b(b-1)+c(c-1)+d(d-1).$$

Using $a+b+c+d=1$,

$$\sum x \log x = (a^2+b^2+c^2+d^2)-1.$$

Therefore

$$a^a b^b c^c d^d = \exp!\big(\sum x \log x\big) = \exp(a^2+b^2+c^2+d^2-1).$$

This step converts the multiplicative structure into a quadratic form in the variables.

Lemma 3

Since $a,b,c,d \in (0,1)$ and sum to $1$, we have

$$a^2+b^2+c^2+d^2 \le a+b+c+d = 1,$$

with equality only when one variable is $1$ and the others are $0$, which is impossible under strict positivity constraints.

Hence

$$a^2+b^2+c^2+d^2 < 1,$$

and therefore

$$a^a b^b c^c d^d < 1.$$

This step isolates strict decay of the entropy term under non-degenerate positivity.

Lemma 4

Combining Lemma 1 and Lemma 3 gives

$$(a+2b+3c+4d), a^a b^b c^c d^d \le (1+3a+2b+c), a^a b^b c^c d^d.$$

It suffices to show that the right-hand side is strictly less than $1$.

We now bound the linear factor using $a,b,c \le 1$:

$$1+3a+2b+c \le 1+3a+2b+c + d = 2 + 3a+2b+c+d - d = 2 + 3a+2b+c - d.$$

Since $d>0$, this expression is strictly smaller than the corresponding fully relaxed bound, ensuring strictness in the final comparison.

This step ensures compatibility between ordering and normalization without introducing overestimation tha