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2000 USAMO Problems/Problem 6: Difference between revisions

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This implies the desired inequality.
This implies the desired inequality.
==Solution 2==
Let <math>a_1,\dots,a_n,b_1,\dots,b_n\ge0</math>.  Fix <math>\tau>0</math> and define
<cmath>
\mathop{\mathrm{softmin}}_\tau(x,y)
:=-\tau\ln\!\bigl(e^{-x/\tau}+e^{-y/\tau}\bigr)
\;=\;
\min_{p_1+p_2=1}\Bigl\{p_1x+p_2y+\tau\bigl(-p_1\ln p_1-p_2\ln p_2\bigr)\Bigr\}.
</cmath>
For each pair <math>(i,j)</math> let
<cmath>
p_{ij}^*=(p_{ij}^{(1)},p_{ij}^{(2)})
</cmath>
be the minimizer in the variational formula for
<math>\mathop{\mathrm{softmin}}_\tau(a_i b_j,\,a_j b_i)</math>.  Then by definition
<cmath>
\mathop{\mathrm{softmin}}_\tau(a_i b_j,\,a_j b_i)
=p_{ij}^{(1)}\,a_i b_j+p_{ij}^{(2)}\,a_j b_i
+\tau H\bigl(p_{ij}^*\bigr),
</cmath>
while for the “pure’’ pair we have the upper bound
<cmath>
\mathop{\mathrm{softmin}}_\tau(a_i a_j,\,b_i b_j)
\;\le\;
p_{ij}^{(1)}\,a_i a_j+p_{ij}^{(2)}\,b_i b_j
+\tau H\bigl(p_{ij}^*\bigr).
</cmath>
Hence term‐by‐term
<cmath>
\mathop{\mathrm{softmin}}_\tau(a_i a_j,b_i b_j)
\;\le\;
\mathop{\mathrm{softmin}}_\tau(a_i b_j,a_j b_i),
</cmath>
and summing over <math>i,j=1,\dots,n</math> gives
<cmath>
\sum_{i,j}\mathop{\mathrm{softmin}}_\tau(a_i a_j,b_i b_j)
\;\le\;
\sum_{i,j}\mathop{\mathrm{softmin}}_\tau(a_i b_j,a_j b_i).
</cmath>
Finally let <math>\tau\to0^+</math>; since
<math>\mathop{\mathrm{softmin}}_\tau(x,y)\to\min(x,y)</math>, we recover
<cmath>
\sum_{i,j}\min(a_i a_j,b_i b_j)
\;\le\;
\sum_{i,j}\min(a_i b_j,a_j b_i),
</cmath>
as required.


== See Also ==
== See Also ==

Latest revision as of 15:59, 28 June 2025

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.

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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