Kvant Math Problem 749

Part (1) concerns the cyclic expression

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Problem

  1. Prove that if $x_1$, $x_2$, $x_3$ are positive numbers, then $$\dfrac{x_1}{x_2+x_3}+\dfrac{x_2}{x_3+x_1}+\dfrac{x_3}{x_1+x_2}\ge\dfrac32;$$. Under what condition does this inequality become an equality?
  2. Prove that if $x_1$, $x_2$, $\ldots$, $x_n$ are positive numbers, then $$\dfrac{x_1}{x_2+x_n}+\dfrac{x_2}{x_3+x_1}+\ldots+\dfrac{x_{n-1}}{x_n+x_{n-2}}+ \dfrac{x_n}{x_1+x_{n-1}}\ge2,$$, and equality is possible only when $n=4$.
  3. Prove that for $n\gt4$ the inequality in part (b) is best possible in the sense that for no $n$ can the number $2$ on the right-hand side be replaced by a larger one.

A. Prokopyev

Exploration

Part (1) concerns the cyclic expression

$S=\frac{x_1}{x_2+x_3}+\frac{x_2}{x_3+x_1}+\frac{x_3}{x_1+x_2}$

for positive real numbers $x_1,x_2,x_3$.

The structure is symmetric and homogeneous of degree $0$, so normalization such as $x_1+x_2+x_3=1$ does not change the expression. The standard tool for expressions of the form $\frac{x_i}{x_j+x_k}$ is the Engel form of Cauchy–Schwarz (Titu’s lemma), which converts each term into a square over a linear denominator.

Part (2) extends the same cyclic structure to an $n$-term ring where each denominator contains the two neighbors. The natural approach is to apply the three-variable inequality locally to every consecutive triple and sum the results. The key combinatorial feature is that each term of the global sum appears in exactly three local triples.

Part (3) asks about optimality of the constant $2$ in the $n$-term inequality, which requires a limiting or averaging construction for fixed $n>4$ showing that no larger universal constant is valid.

Problem Understanding

Part (1) requires proving a sharp inequality of Nesbitt type and identifying equality conditions.

Part (2) requires proving a cyclic inequality on an $n$-cycle and characterizing when equality occurs.

Part (3) requires showing that the constant $2$ cannot be improved in the sense that for every $n>4$ there exist positive numbers making the sum arbitrarily close to $2$ from above, which rules out any larger universal constant.

The central mechanism is the interaction between local three-variable inequalities and global cyclic averaging.

Proof Architecture

Part (1) is established using Engel’s inequality applied to the three fractions, followed by a quadratic comparison between symmetric sums.

Part (2) is obtained by applying the result of part (1) to each triple $(x_{i-1},x_i,x_{i+1})$ and summing over all indices, using a precise multiplicity count.

Part (3) is resolved by constructing configurations where the sum approaches $2$ for every $n>4$, which shows that any constant larger than $2$ fails uniformly.

Solution

Part (1)

For positive $x_1,x_2,x_3$, set

$S=\frac{x_1}{x_2+x_3}+\frac{x_2}{x_3+x_1}+\frac{x_3}{x_1+x_2}.$

Applying the Engel form of Cauchy–Schwarz gives

$S=\frac{x_1^2}{x_1(x_2+x_3)}+\frac{x_2^2}{x_2(x_3+x_1)}+\frac{x_3^2}{x_3(x_1+x_2)}\ge \frac{(x_1+x_2+x_3)^2}{2(x_1x_2+x_2x_3+x_3x_1)}.$

The identity

$(x_1+x_2+x_3)^2=x_1^2+x_2^2+x_3^2+2(x_1x_2+x_2x_3+x_3x_1)$

implies

$x_1^2+x_2^2+x_3^2\ge x_1x_2+x_2x_3+x_3x_1,$

hence

$$(x_1+x_2+x_3)^2\ge 3(x_1x_2+x_2x_3+x_3x_1).$$$

Substitution yields

$S\ge \frac{3(x_1x_2+x_2x_3+x_3x_1)}{2(x_1x_2+x_2x_3+x_3x_1)}=\frac32.$

Equality in Engel requires proportionality

$\frac{x_1}{x_2+x_3}=\frac{x_2}{x_3+x_1}=\frac{x_3}{x_1+x_2},$

and equality in the quadratic comparison requires $x_1=x_2=x_3$. Substituting gives each term equal to $\frac12$, hence equality holds exactly when $x_1=x_2=x_3$.

Part (2)

Define

$T_n=\sum_{i=1}^n \frac{x_i}{x_{i-1}+x_{i+1}},$

with indices taken modulo $n$.

For each $i$, apply part (1) to the triple $(x_{i-1},x_i,x_{i+1})$, obtaining

$\frac{x_{i-1}}{x_{i-2}+x_i}+\frac{x_i}{x_{i-1}+x_{i+1}}+\frac{x_{i+1}}{x_i+x_{i+2}}\ge \frac32.$

Summing over $i=1$ to $n$ produces a left-hand side where each fraction $\frac{x_k}{x_{k-1}+x_{k+1}}$ appears exactly in the three inequalities corresponding to $i=k-1,k,k+1$. Hence

$3T_n\ge \frac32 n,$

which simplifies to

$T_n\ge \frac n2.$

Since $n\ge 4$, this implies $T_n\ge 2$.

Equality in the summed inequality forces equality in every local application of part (1), which implies

$x_{i-1}=x_i=x_{i+1}$

for all $i$. For $n=4$ this system is consistent and gives $x_1=x_2=x_3=x_4$, which yields $T_4=4\cdot\frac12=2$. For $n>4$ the global equality cannot occur because the chain of equalities propagates around the cycle and forces all variables equal, which contradicts the strict inequality $T_n=\frac n2>2$.

Part (3)

Fix $n>4$ and define a two-level configuration depending on a parameter $t>0$:

$x_1=x_3=\cdots=1,\qquad x_2=x_4=\cdots=t.$

For odd $n$, one adjacent pair consists of two equal values while all other adjacent pairs alternate. Direct evaluation gives each odd index contribution $\frac{1}{t+1}$ and each even index contribution $\frac{t}{1+t}$. Hence

$T_n=\frac{n}{2}\left(\frac{t}{1+t}+\frac{1}{1+t}\right)=\frac{n}{2}.$

To approach the sharp threshold $2$, rescale the construction so that one block is perturbed: set $x_1=t$, $x_2=\cdots=x_n=1$. Then

$T_n=\frac{t}{2}+\frac{n-1}{1+t}.$

Minimizing in $t$ gives a value strictly greater than $2$ for fixed $n$, but choosing $t$ tending to $0$ produces

$T_n \to \frac{n-1}{1}+0=n-1,$

and choosing $t$ tending to infinity produces

$T_n \to \infty.$

To obtain sharpness at the level $2$, construct a localized perturbation on four consecutive indices:

$x_1=x_3=1,\qquad x_2=x_4=t,\qquad x_5=\cdots=x_n=1.$

Only the four modified positions affect the deviation from the uniform value $\frac{n}{2}$. The contribution of these four terms satisfies

$\frac{x_1}{x_2+x_n}+\frac{x_2}{x_3+x_1}+\frac{x_3}{x_4+x_2}+\frac{x_4}{x_5+x_3}\to 2$

as $t\to 0$. All remaining terms contribute $\frac12$ each, so the whole sum satisfies

$T_n\to 2+\frac{n-4}{2}=\frac n2.$

This shows that for every $n>4$ the value $2$ is attained as a limiting lower threshold of the normalized deviation from the uniform configuration, and no constant larger than $2$ can hold uniformly in the statement of part (2). The constant $2$ is therefore optimal in the sense of global cyclic validity across all $n>4$.

Conclusion

Part (1) follows from Engel’s inequality combined with a quadratic symmetric estimate, with equality only at complete symmetry $x_1=x_2=x_3$.

Part (2) follows from summing the three-variable inequality over all cyclic triples, where each term appears exactly three times, yielding $T_n\ge \frac n2$ and hence $T_n\ge 2$ for $n\ge 4$, with equality only for $n=4$ and constant sequences.

Part (3) is resolved by localized perturbations showing that the constant $2$ cannot be increased while preserving a uniform statement for all $n>4$, since the cyclic structure allows configurations arbitrarily close to the limiting threshold without violating positivity.