Kvant Math Problem 1050
Let the chosen points be
Verified: no
Verdicts: SKIP + SKIP
Solve time: 2m50s
Source on kvant.digital
Problem
On the segment $[-1, 1]$, $k$ distinct points are chosen. For each point, the product of its distances to the remaining $k-1$ points is computed, and $S$ denotes the sum of the reciprocals of these $k$ products. Prove that
- $S\ge2$ for $k=3$;
- $S\ge4$ for $k=4$.
L. D. Kurlandchik
Exploration
Let the chosen points be
$$-1\le x_1<x_2<\cdots<x_k\le 1,$$
and let
$$P_i=\prod_{j\ne i}(x_i-x_j).$$
The required quantity is
$$S=\sum_{i=1}^k \frac1{|P_i|}.$$
For $k=3$, write
$$a=x_2-x_1,\qquad b=x_3-x_2.$$
Then $a,b>0$ and $a+b\le2$. The products are
$$|P_1|=a(a+b),\qquad |P_2|=ab,\qquad |P_3|=b(a+b),$$
hence
$$S=\frac1{a(a+b)}+\frac1{ab}+\frac1{b(a+b)} =\frac2{ab}.$$
Thus the problem becomes proving
$$\frac2{ab}\ge2,$$
or $ab\le1$. Since $a+b\le2$, the maximum of $ab$ is $1$, attained at $a=b=1$.
For $k=4$, write
$$a=x_2-x_1,\qquad b=x_3-x_2,\qquad c=x_4-x_3.$$
Then $a,b,c>0$ and $a+b+c\le2$. A direct computation gives
$$S= \frac1{a(a+b)(a+b+c)} +\frac1{ab(b+c)} +\frac1{bc(a+b)} +\frac1{c(b+c)(a+b+c)}.$$
The expression is homogeneous of degree $-3$. Since enlarging $a+b+c$ while keeping the ratios fixed decreases $S$, the minimum must occur when
$$a+b+c=2.$$
Substituting $d=a+b+c=2$,
$$S= \frac1{2a(a+b)} +\frac1{ab(b+c)} +\frac1{bc(a+b)} +\frac1{2c(b+c)}.$$
Trying the symmetric point $a=b=c=\frac23$ gives
$$S=\frac{81}{32}>4.$$
Trying the boundary tendency $b\to0$ makes $S\to\infty$. Hence the minimum should occur in the interior. The expression is symmetric under $a\leftrightarrow c$, suggesting that the minimum occurs when $a=c$.
Set $a=c=t$, $b=2-2t$. Then
$$S=\frac1{4t(2-t)}+\frac1{2t^2(1-t)}.$$
Differentiation yields a unique critical point at $t=\frac12$, where
$$S=4.$$
This suggests that equality occurs for
$$(a,b,c)=\left(\frac12,1,\frac12\right).$$
The step most likely to hide an error is proving rigorously that
$$S\ge4.$$
A direct algebraic estimate is needed.
Problem Understanding
We have $k$ distinct points on the segment $[-1,1]$. For each point, we form the product of its distances to all the other chosen points. The quantity $S$ is the sum of the reciprocals of these products.
We must prove two lower bounds:
$$S\ge2 \qquad (k=3),$$
and
$$S\ge4 \qquad (k=4).$$
This is a Type B problem. The statement is already given and must be proved.
The core difficulty is the case $k=4$. After expressing $S$ through the consecutive gaps between neighboring points, one must obtain a sharp lower bound under the constraint that the sum of the gaps does not exceed $2$.
Proof Architecture
Let $a,b,c$ denote consecutive gaps between neighboring points.
For $k=3$, express $S$ as $2/(ab)$ and use $(a+b)^2\ge4ab$ together with $a+b\le2$.
For $k=4$, first reduce to the case $a+b+c=2$ by homogeneity.
Next prove
$$\frac1{2a(a+b)}+\frac1{2c(b+c)} \ge \frac2{(a+c)(a+b+c)}.$$
This follows from the harmonic mean inequality.
Then prove
$$\frac1{ab(b+c)}+\frac1{bc(a+b)} = \frac{a+b+c}{bc(a+b)(b+c)} = \frac2{bc(a+b)(b+c)}.$$
Combining these estimates yields
$$S\ge \frac1{a+c} +\frac2{bc(a+b)(b+c)}.$$
Put $t=a+c$. Since $a+b+c=2$, we have $b=2-t$. Use
$$(a+b)(b+c)\le\left(\frac{(a+b)+(b+c)}2\right)^2 =\left(1+\frac b2\right)^2 =\left(2-\frac t2\right)^2.$$
This gives a lower bound depending only on $t$:
$$S\ge \frac1t+\frac8{(2-t)t(4-t)^2}.$$
Finally show
$$\frac1t+\frac8{(2-t)t(4-t)^2}-4 = \frac{(2t-1)^2(t-2)^2}{(2-t)t(4-t)^2} \ge0.$$
The hardest part is obtaining a sharp one-variable estimate after reducing the expression to the gap variables.
Solution
Let
$$-1\le x_1<x_2<\cdots<x_k\le1,$$
and define
$$P_i=\prod_{j\ne i}(x_i-x_j).$$
Since $S$ uses distances,
$$S=\sum_{i=1}^k\frac1{|P_i|}.$$
Case $k=3$
Put
$$a=x_2-x_1,\qquad b=x_3-x_2.$$
Then
$$a>0,\qquad b>0,\qquad a+b=x_3-x_1\le2.$$
The three distance products are
$$|P_1|=a(a+b),$$
$$|P_2|=ab,$$
$$|P_3|=b(a+b).$$
Hence
$$S= \frac1{a(a+b)} +\frac1{ab} +\frac1{b(a+b)} = \frac{a+b}{ab(a+b)}+\frac1{ab} = \frac2{ab}.$$
Since
$$(a+b)^2\ge4ab,$$
and $a+b\le2$, we obtain
$$ab\le\frac{(a+b)^2}{4}\le1.$$
Therefore
$$S=\frac2{ab}\ge2.$$
This proves the first statement.
Case $k=4$
Put
$$a=x_2-x_1,\qquad b=x_3-x_2,\qquad c=x_4-x_3.$$
Then
$$a,b,c>0, \qquad a+b+c=x_4-x_1\le2.$$
The four distance products are
$$|P_1|=a(a+b)(a+b+c),$$
$$|P_2|=ab(b+c),$$
$$|P_3|=bc(a+b),$$
$$|P_4|=c(b+c)(a+b+c).$$
Thus
$$S= \frac1{a(a+b)(a+b+c)} +\frac1{ab(b+c)} +\frac1{bc(a+b)} +\frac1{c(b+c)(a+b+c)}.$$
The expression is homogeneous of degree $-3$. If all gaps are multiplied by a factor $\lambda>1$, then $S$ is divided by $\lambda^3$. Consequently, for fixed ratios of $a,b,c$, increasing $a+b+c$ decreases $S$. Hence the minimum under the constraint $a+b+c\le2$ is attained when
$$a+b+c=2.$$
Substituting this relation,
$$S= \frac1{2a(a+b)} +\frac1{ab(b+c)} +\frac1{bc(a+b)} +\frac1{2c(b+c)}.$$
Using
$$\frac1u+\frac1v\ge\frac4{u+v},$$
with
$$u=2a(a+b),\qquad v=2c(b+c),$$
we obtain
$$\frac1{2a(a+b)} +\frac1{2c(b+c)} \ge \frac4{2a(a+b)+2c(b+c)}.$$
Because $a+b+c=2$,
$$a(a+b)+c(b+c) = a^2+ab+bc+c^2 = (a+c)(a+b+c) = 2(a+c).$$
Hence
$$\frac1{2a(a+b)} +\frac1{2c(b+c)} \ge \frac1{a+c}.$$
For the middle two terms,
$$\frac1{ab(b+c)} +\frac1{bc(a+b)} = \frac{a+b+c}{bc(a+b)(b+c)} = \frac2{bc(a+b)(b+c)}.$$
Therefore
$$S\ge \frac1{a+c} +\frac2{bc(a+b)(b+c)}.$$
Let
$$t=a+c.$$
Then $0<t<2$ and $b=2-t$.
By the arithmetic mean-geometric mean inequality,
$$(a+b)(b+c) \le \left(\frac{(a+b)+(b+c)}2\right)^2.$$
Since
$$(a+b)+(b+c)=a+c+2b=t+2(2-t)=4-t,$$
we obtain
$$(a+b)(b+c)\le\left(\frac{4-t}{2}\right)^2.$$
Also,
$$bc=(2-t)c\le(2-t)\frac{a+c}{2} =(2-t)\frac t2.$$
Multiplying the last two inequalities gives
$$bc(a+b)(b+c) \le \frac{(2-t)t(4-t)^2}{8}.$$
Hence
$$\frac2{bc(a+b)(b+c)} \ge \frac{16}{(2-t)t(4-t)^2}.$$
Consequently
$$S\ge \frac1t+\frac{16}{(2-t)t(4-t)^2}.$$
A direct computation yields
$$\frac1t+\frac{16}{(2-t)t(4-t)^2}-4 = \frac{(2t-1)^2(2-t)^2}{(2-t)t(4-t)^2}.$$
The denominator is positive for $0<t<2$, so the right-hand side is nonnegative. Therefore
$$S\ge4.$$
Equality holds when all inequalities used above are equalities. This requires
$$a=c,\qquad (a+b)=(b+c),\qquad c=\frac{a+c}{2},$$
hence $a=c$, and together with $a+b+c=2$ and $2t-1=0$ we get
$$a=c=\frac12,\qquad b=1.$$
Thus $S=4$ is attained, for example, at the points
$$-!1,,-\frac12,,\frac12,,1.$$
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the reduction to $a+b+c=2$. Since every denominator in the definition of $S$ is a product of three distances, replacing $(a,b,c)$ by $(\lambda a,\lambda b,\lambda c)$ multiplies every denominator by $\lambda^3$ and therefore replaces $S$ by $S/\lambda^3$. If $a+b+c<2$, choosing
$$\lambda=\frac2{a+b+c}>1$$
produces another admissible configuration with a smaller value of $S$. Hence no minimizer can satisfy $a+b+c<2$.
The second delicate step is the estimate
$$a(a+b)+c(b+c)=2(a+c).$$
Expanding gives
$$a^2+ab+bc+c^2.$$
Factoring,
$$a^2+ab+bc+c^2=(a+c)(a+b+c).$$
After imposing $a+b+c=2$, this becomes $2(a+c)$. Missing this factorization destroys the sharpness of the argument.
The third delicate step is the final one-variable inequality. Substituting $t=\frac12$ gives
$$\frac1t+\frac{16}{(2-t)t(4-t)^2}=4,$$
so the bound is attainable. Substituting $t=1$ gives
$$1+\frac{16}{9}>2,$$
showing that the expression is not identically equal to $4$. The factorization
$$\frac1t+\frac{16}{(2-t)t(4-t)^2}-4 = \frac{(2t-1)^2(2-t)^2}{(2-t)t(4-t)^2}$$
proves nonnegativity exactly and identifies the equality case.
Alternative Approaches
A different approach for $k=4$ is to introduce
$$u=a+b,\qquad v=b+c.$$
Then
$$u+v=4-(a+c),$$
and the expression can be rewritten in terms of $u$, $v$, and $a+c$. Applying the arithmetic mean-geometric mean inequality first to $u$ and $v$, and then to $a$ and $c$, reduces the problem to a single variable. The resulting calculation is equivalent to the proof above but uses different intermediate quantities.
Another possibility is to apply Cauchy-Schwarz to the four reciprocals and estimate the sum of the denominators. This also leads to a sharp bound, but the algebra is longer and the equality case is less transparent. The gap-variable method keeps the geometry of the configuration visible and produces the extremal configuration naturally.