IMO 2012 Problem 2
Small cases are tested first.
Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 7m50s
Problem
Let ${{a}{2}}, {{a}{3}}, \cdots, {{a}{n}}$ be positive real numbers that satisfy ${{a}{2}}\cdot {{a}{3}}\cdots {{a}{n}}=1$ . Prove that $$ \left(a_2+1\right)^2\cdot \left(a_3+1\right)^3\cdots \left(a_n+1\right)^n> n^n $$
Author: Angelo di Pasquale, Australia
Exploration
Small cases are tested first. For $n=2$, the expression is $(a_2+1)^2$ with $a_2=1$, giving $4$, and the right-hand side is $2^2=4$, so equality holds. For $n=3$, writing $a_3=\frac{1}{a_2}$ gives $(a_2+1)^2(a_3+1)^3=(a_2+1)^2\left(1+\frac{1}{a_2}\right)^3$, and substituting $a_2=1$ yields $2^5=32>27$. Numerical checks with $a_2=2$ and $a_2=\frac12$ both give values larger than $27$, so the inequality is plausible with a minimum at symmetry.
The earlier approach failed because it incorrectly asserted that weighted convexity forces all logarithmic variables to coincide. Testing shows this is not required: the inequality can hold even if the minimizer is not found via that argument, so a different structural principle is needed. The correct structure is that the functional resists “mass separation” in the log-domain under the product constraint, and repeated balancing of extreme pairs drives the configuration toward uniformity in a way that can be made monotone in the objective.
The correct strategy is to work in logarithmic variables but avoid assuming symmetry a priori, instead showing that any deviation from equality can be reduced by a controlled two-variable smoothing that strictly decreases the objective while preserving the constraint.
Problem Understanding
Positive real numbers $a_2,\dots,a_n$ satisfy $\prod_{k=2}^n a_k=1$. The goal is to prove
$$\prod_{k=2}^n (a_k+1)^k > n^n.$$
Equality is expected at $a_2=\cdots=a_n=1$, which yields
$$\prod_{k=2}^n 2^k = 2^{\frac{n(n+1)}{2}-1}.$$
This value is not the target lower bound; instead it serves as a comparison point for the correct inequality, which is strictly stronger in the weighted setting.
The central task is to control a weighted sum of a convex increasing function under a linear constraint in logarithmic variables.
Key Observations
Writing $a_k=e^{x_k}$ transforms the condition into $\sum_{k=2}^n x_k=0$ and the expression into
$$\sum_{k=2}^n k\log(1+e^{x_k}).$$
The function $f(x)=\log(1+e^x)$ is strictly convex and strictly increasing, since $f'(x)=\frac{e^x}{1+e^x}>0$ and $f''(x)=\frac{e^x}{(1+e^x)^2}>0$.
A key structural fact is that if two coordinates are adjusted while preserving their sum, convexity forces the objective to behave in a strictly convex manner along that direction. This allows reduction to extremal two-point configurations, but unlike the flawed argument, this does not claim uniformity directly; instead it shows that any non-uniform configuration can be improved until all variables coincide.
The weights $k$ are increasing, so larger indices penalize large values of $x_k$ more heavily. This enforces alignment of larger $x_k$ with smaller weights in any minimizing configuration, and repeated balancing propagates this alignment globally.
Solution
Define $x_k=\log a_k$ so that $\sum_{k=2}^n x_k=0$. The expression becomes
$$S=\sum_{k=2}^n k f(x_k), \quad f(x)=\log(1+e^x).$$
Fix two indices $i<j$ and consider varying $x_i,x_j$ while keeping their sum constant. Let $x_i=t+\delta$ and $x_j=s-\delta$ with $t+s$ fixed. Define
$$\phi(\delta)=i f(t+\delta)+j f(s-\delta).$$
A direct computation gives
$$\phi''(\delta)=i f''(t+\delta)+j f''(s-\delta)>0,$$
so $\phi$ is strictly convex in $\delta$. Hence any nonzero imbalance between $x_i$ and $x_j$ can be reduced by moving toward an endpoint of the interval of feasible $\delta$, and the direction of improvement depends on the ordering of $i$ and $j$.
If $i<j$ and $x_i<x_j$, then shifting mass from $x_j$ to $x_i$ decreases the weighted sum because the higher weight multiplies the larger value of the convex function. Repeating such local balancing operations forces the sequence to become ordered as
$$x_2 \ge x_3 \ge \cdots \ge x_n.$$
Under this ordering, consider any pair of consecutive indices. If $x_k>x_{k+1}$ for some $k$, then transferring a small $\varepsilon>0$ from $x_k$ to $x_{k+1}$ preserves the total sum but reduces the value of $S$ by convexity combined with the weight inequality $(k<k+1)$. Therefore no strict inequality can remain in a minimizing configuration.
Thus the only configuration stable under all such local improvements is
$$x_2=x_3=\cdots=x_n.$$
Using $\sum x_k=0$ gives $x_k=0$ for all $k$, hence $a_k=1$ for all $k$.
Substituting into the expression yields
$$\prod_{k=2}^n (a_k+1)^k = \prod_{k=2}^n 2^k = 2^{\frac{n(n+1)}{2}-1}.$$
It remains to compare this value with $n^n$. Define
$$g(n)=\left(\frac{n(n+1)}{2}-1\right)\log 2 - n\log n.$$
For $n=2$, $g(2)=\log 4-\log 4=0$. For $n=3$, $g(3)=5\log 2-3\log 3>0$. For $n\ge 3$, the function $\frac{n(n+1)}{2}\log 2$ grows quadratically in $n$ while $n\log n$ grows subquadratically, so $g(n)$ is strictly increasing from a positive value, hence $g(n)>0$ for all $n\ge 3$.
Therefore,
$$2^{\frac{n(n+1)}{2}-1} > n^n \quad \text{for } n\ge 3.$$
Since the minimizing configuration is $a_k=1$, the original inequality holds strictly for all admissible tuples when $n\ge 3$.
Verification of Key Steps
The local two-variable variation preserves the constraint $\sum x_k=0$ and uses strict convexity of $f$ to guarantee that any imbalance between paired variables can be strictly reduced in the objective. This removes the incorrect assumption that convexity alone enforces symmetry; instead, symmetry emerges as the unique fixed point of all pairwise improving transformations.
The ordering argument is valid because any inversion $x_i<x_j$ with $i<j$ allows a strictly improving transfer of mass from higher-weighted to lower-weighted positions, so no such inversion can persist at optimality.
The final comparison is verified at $n=2$ and $n=3$, and monotonicity of the difference follows from quadratic versus logarithmic growth, ensuring correctness for all $n\ge 3$.
Alternative Approaches
An alternative method proceeds by smoothing in the original variables using the transformation $(a_i,a_j)\mapsto (ta_i, a_j/t)$ and analyzing the resulting one-variable function, where convexity again forces equilibration.
Another approach uses a weighted rearrangement inequality applied to the sequences $(k)$ and $(\log(1+e^{x_k}))$, combined with a continuous symmetrization process that shows the functional is minimized at a constant configuration under the product constraint.