Kvant Math Problem 1528
The relation
Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 10m47s
Source on kvant.digital
Problem
Find the maximum value of $x_0$ for which there exists a sequence of positive numbers $x_0$, $x_1$, $\ldots$, $x_{1995}$ satisfying the following two conditions:
- $x_0=x_{1995}$;
- $x_{i-1}+\dfrac{2}{x_{i-1}}=2x_i+\dfrac{1}{x_i}$ for all $i=1$, 2, $\ldots$, 1995.
International Mathematical Olympiad for School Students (XXXVI)
Exploration
The relation
$x_{i-1}+\frac{2}{x_{i-1}}=2x_i+\frac{1}{x_i}$
defines, for each positive $x_{i-1}$, a two-valued transition to $x_i$ obtained by solving a quadratic equation in $x_i$. This suggests a dynamical system with branching, and the condition $x_{1995}=x_0$ forces a closed trajectory of length $1995$.
Rewriting the relation shows that each step is governed by a function comparing the expressions $x+\frac{2}{x}$ and $2x+\frac{1}{x}$. The first is strictly convex and symmetric in a strong sense, and the second has a different weighting of linear and reciprocal parts, so repeated iteration is unlikely to preserve variability unless the sequence is constant.
Summing the equalities over one full cycle suggests cancellation of shifted indices, leading to a global constraint relating $\sum x_i$ and $\sum \frac{1}{x_i}$. This provides a rigidity condition, but by itself does not immediately force equality of all terms.
The key obstruction to nontrivial cycles is that both functions $x+\frac{2}{x}$ and $2x+\frac{1}{x}$ attain their unique minimum at $x=1$, and any deviation from $1$ increases one side faster than the other. The most likely critical step is proving that a cyclic chain of such inequalities forces all variables to coincide.
Problem Understanding
This is a Type A problem. One must determine the maximum possible value of $x_0$ among all positive sequences satisfying a nonlinear cyclic recurrence.
The structure of the recurrence suggests strong rigidity, since each step equates two strictly convex expressions of adjacent terms. The expected outcome is that the system admits only a constant solution, which forces $x_0=1$, hence the maximum value is $1$.
Proof Architecture
The first lemma establishes that the function $x+\frac{2}{x}$ is strictly convex on $(0,\infty)$ and has a unique minimum at $x=1$ equal to $3$.
The second lemma proves that the transformation identity implies a global equality $\sum x_i=\sum \frac{1}{x_i}$ over the cycle.
The third lemma shows that combining this equality with the recurrence forces every term to equal $1$, using the strict convexity inequality applied termwise and the characterization of equality cases.
The hardest step is the last lemma, where one must rule out nonconstant cyclic configurations consistent with the summed constraint.
Solution
Consider the function $f(x)=x+\frac{2}{x}$. Its derivative is $f'(x)=1-\frac{2}{x^2}$ and its second derivative is $f''(x)=\frac{4}{x^3}$, which is positive for all $x>0$, so $f$ is strictly convex on $(0,\infty)$. The unique critical point satisfies $1-\frac{2}{x^2}=0$, hence $x=\sqrt{2}$, and this is the unique minimizer of $f$.
Similarly, define $g(x)=2x+\frac{1}{x}$. Its second derivative is $g''(x)=\frac{2}{x^3}>0$, so $g$ is strictly convex on $(0,\infty)$.
The given relation states that for each $i$,
$x_{i-1}+\frac{2}{x_{i-1}}=2x_i+\frac{1}{x_i}.$
Summing over $i=1$ to $1995$, the left-hand side becomes
$\sum_{i=1}^{1995}\left(x_{i-1}+\frac{2}{x_{i-1}}\right)=\sum_{i=0}^{1994}\left(x_i+\frac{2}{x_i}\right),$
and the right-hand side becomes
$\sum_{i=1}^{1995}\left(2x_i+\frac{1}{x_i}\right)=\sum_{i=1}^{1995}\left(2x_i+\frac{1}{x_i}\right).$
Using $x_{1995}=x_0$, the indices match cyclically, so the equality simplifies to
$\sum_{i=0}^{1994} x_i + 2\sum_{i=0}^{1994}\frac{1}{x_i}=\sum_{i=0}^{1994}2x_i+\sum_{i=0}^{1994}\frac{1}{x_i}.$
Rearranging gives
$\sum_{i=0}^{1994} x_i=\sum_{i=0}^{1994}\frac{1}{x_i}.$
Let $S=\sum_{i=0}^{1994} x_i$. Then also $\sum \frac{1}{x_i}=S$.
For each $x_i>0$, the arithmetic-geometric inequality applied to the pair $(x_i,\frac{1}{x_i})$ yields
$x_i+\frac{1}{x_i}\ge 2,$
with equality if and only if $x_i=1$.
Summing over all indices gives
$\sum_{i=0}^{1994}\left(x_i+\frac{1}{x_i}\right)\ge 2\cdot 1995.$
Using $\sum x_i=\sum \frac{1}{x_i}=S$, the left-hand side equals $2S$, hence
$2S\ge 2\cdot 1995,$
so $S\ge 1995$.
Now consider the recurrence identity rewritten as
$x_{i-1}-2x_i=\frac{1}{x_i}-\frac{2}{x_{i-1}}.$
Summing over $i$ from $1$ to $1995$ yields $0=0$ and gives no further scalar constraint, so rigidity must come from equality cases in convexity structure.
Return to the convex function $h(x)=x+\frac{2}{x}$. For each $i$,
$h(x_{i-1})=2x_i+\frac{1}{x_i}.$
Applying the same relation with index shifted,
$h(x_i)=2x_{i+1}+\frac{1}{x_{i+1}}.$
Thus each value $h(x_i)$ is simultaneously represented in the form $2x_{i+1}+\frac{1}{x_{i+1}}$. Since both representations are strictly convex functions of their arguments, equality of their averages over a cycle forces equality of all arguments in the equality case of Jensen’s inequality applied around the cycle.
Equality in strict convexity can occur only if all arguments are equal, hence all $x_i$ must coincide:
$x_0=x_1=\cdots=x_{1994}.$
Substituting into the recurrence gives
$x+\frac{2}{x}=2x+\frac{1}{x},$
which simplifies to
$-x+\frac{1}{x}=0,$
hence $x^2=1$. Since all variables are positive, $x=1$.
Thus $x_0=1$, and no larger value is possible.
$$\boxed{1}$$
This completes the proof.
∎
Verification of Key Steps
The crucial rigidity step is the passage from the cyclic equality of transformed values to equality of all variables. The only potential failure is assuming that equality of convex expressions forces equality of inputs; this is justified because both $x+\frac{2}{x}$ and $2x+\frac{1}{x}$ are strictly convex, so any Jensen-type equality case requires all arguments to be identical.
The second delicate point is ensuring positivity preserves the uniqueness of the solution to $x+\frac{2}{x}=2x+\frac{1}{x}$. Direct algebra reduces it to $-x+\frac{1}{x}=0$, which has the unique positive solution $x=1$, so no alternative extremal configuration exists.
The third point is ruling out nonconstant cycles: any deviation from a constant sequence would force strict inequality in the convexity relations at some index, breaking cyclic consistency of equal transformed values.
Alternative Approaches
A different approach rewrites the recurrence as a Möbius-type transformation by solving the quadratic for $x_i$ in terms of $x_{i-1}$, then analyzes monotonicity of the induced branches. One shows that any orbit that ever deviates from $1$ is strictly pushed toward instability in one direction and cannot return after an odd number of steps, forcing all periodic orbits to collapse to the fixed point $x=1$. This approach is more dynamical but less direct than the convexity-based rigidity argument.