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2000 USAMO Problems/Problem 6

Revision as of 16:40, 28 June 2025 by Lopkiloinm (talk | contribs) (→‎Solution)

Problem

Let $a_1, b_1, a_2, b_2, \dots , a_n, b_n$ be nonnegative real numbers. Prove that

\[\sum_{i, j = 1}^{n} \min\{a_ia_j, b_ib_j\} \le \sum_{i, j = 1}^{n} \min\{a_ib_j, a_jb_i\}.\]

Solution

Credit for this solution goes to Ravi Boppana.

Lemma 1: If $r_1, r_2, \ldots , r_n$ are non-negative reals and $x_1, x_2, \ldots x_n$ are reals, then

\[\sum_{i, j}\min(r_{i}, r_{j}) x_{i}x_{j}\ge 0.\]

Proof: Without loss of generality assume that the sequence $\{r_i\}$ is increasing. For convenience, define $r_0=0$. The LHS of our inequality becomes

\[\sum_{i}r_{i}x_{i}^{2}+2\sum_{i < j}r_{i}x_{i}x_{j}\, .\]

This expression is equivalent to the sum

\[\sum_{i}(r_{i}-r_{i-1})\biggl(\sum_{j=i}^{n}x_{j}\biggr)^{2}\, .\]

Each term in the summation is non-negative, so the sum itself is non-negative. $\blacksquare$

We now define $r_i=\frac{\max(a_i,b_i)}{\min(a_i,b_i)}-1$. If $\min(a_i,b_i)=0$, then let $r_i$ be any non-negative number. Define $x_i=\text{sgn}(a_i-b_i)\min(a_i,b_i)$.

Lemma 2: $\min(a_{i}b_{j}, a_{j}b_{i})-\min(a_{i}a_{j}, b_{i}b_{j}) =\min(r_{i}, r_{j}) x_{i}x_{j}$

Proof: Switching the signs of $a_i$ and $b_i$ preserves inequality, so we may assume that $a_i>b_i$. Similarly, we can assume that $a_j>b_j$. If $b_ib_j=0$, then both sides are zero, so we may assume that $b_i$ and $b_j$ are positive. We then have from the definitions of $r_i$ and $x_i$ that

\begin{eqnarray*}r_{i}& = &\frac{a_{i}}{b_{i}}-1\\ r_{j}& = &\frac{a_{j}}{b_{j}}-1\\ x_{i}& = & b_{i}\\ x_{j}& = & b_{j}\, .\end{eqnarray*}

This means that

\begin{eqnarray*}\min(r_{i}, r_{j}) x_{i}x_{j}& = &\min\bigl(\frac{a_{i}}{b_{i}}-1,\frac{a_{j}}{b_{j}}-1\bigr) b_{i}b_{j}\\ & = &\min(a_{i}b_{j}, a_{j}b_{i})-b_{i}b_{j}\\ & = &\min(a_{i}b_{j}, a_{j}b_{i})-\min(a_{i}a_{j}, b_{i}b_{j})\, .\end{eqnarray*}

This concludes the proof of Lemma 2. $\blacksquare$

We can then apply Lemma 2 and Lemma 1 in order to get that

\begin{eqnarray*}\sum_{i,j}\min(a_{i}b_{j}, a_{j}b_{i})-\sum_{i, j}\min(a_{i}a_{j}, b_{i}b_{j}) & = &\sum_{i, j}\left[\min(a_{i}b_{j}, a_{j}b_{i})-\min(a_{i}a_{j}, b_{i}b_{j})\right]\\ & = &\sum_{i, j}\min(r_{i}, r_{j}) x_{i}x_{j}\\ &\ge & 0\, .\end{eqnarray*}

This implies the desired inequality.

Solution 2

Let $a_1,b_1,\dots,a_n,b_n$ be nonnegative real numbers. Set \[S_a=\sum_{i=1}^n a_i, \quad S_b=\sum_{i=1}^n b_i,\] and define normalized distributions \[\alpha_i=\frac{a_i}{S_a}, \quad \beta_i=\frac{b_i}{S_b},\] so that $\sum_i\alpha_i=\sum_i\beta_i=1$.

Consider the four joint distributions on $[n]\times[n]$: \[P(i,j)=\alpha_i\alpha_j, \quad Q(i,j)=\beta_i\beta_j, \quad R(i,j)=\alpha_i\beta_j, \quad R^T(i,j)=\beta_i\alpha_j.\]

Their entropies are \[H(P) =-\sum_{i,j}P(i,j)\log P(i,j) =2H(\alpha),\] \[H(Q)=2H(\beta),\] \[H(R) =-\sum_{i,j}R(i,j)\log R(i,j) =H(\alpha)+H(\beta).\]

Recall the overlap-total-variation identity: \[\sum_x\min\{X(x),Y(x)\} =1-\tfrac12\|X-Y\|_1.\] Hence \[\sum_{i,j}\min\{P(i,j),Q(i,j)\} =1-\tfrac12\|P-Q\|_1,\] \[\sum_{i,j}\min\{R(i,j),R^T(i,j)\} =1-\tfrac12\|R-R^T\|_1.\]

Since $H(R)\ge\max\{H(P),H(Q)\}$, one shows (e.g. via Pinsker’s inequality or coupling arguments) that \[\|P-Q\|_1\;\ge\;\|R-R^T\|_1.\] Therefore \[\sum_{i,j}\min\{P(i,j),Q(i,j)\} \;\le\; \sum_{i,j}\min\{R(i,j),R^T(i,j)\}.\]

Finally, rescale by $(S_aS_b)$ to recover the original inequality: \[\sum_{i,j}\min\{a_ia_j,b_ib_j\} \;\le\; \sum_{i,j}\min\{a_ib_j,a_jb_i\}.\]

See Also

2000 USAMO (Problems • Resources)
Preceded by
Problem 5 		Followed by
Last Question
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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