IMO 2013 Problem 5

Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 5m21s

Problem

Let $\mathbb Q_{>0}$ be the set of all positive rational numbers. Let $f:\mathbb Q_{>0}\to\mathbb R$ be a function satisfying the following three conditions:

(i) for all $x,y\in\mathbb Q_{>0}$, we have $f(x)f(y)\geq f(xy)$; (ii) for all $x,y\in\mathbb Q_{>0}$, we have $f(x+y)\geq f(x)+f(y)$; (iii) there exists a rational number $a>1$ such that $f(a)=a$.

Prove that $f(x)=x$ for all $x\in\mathbb Q_{>0}$.

Exploration

The functional inequalities suggest a comparison between multiplicative and additive behavior, together with a normalization condition at a single rational point $a>1$ where $f(a)=a$. The structure resembles a rigidity problem: a function simultaneously “super-additive” in the sense of (ii) and “sub-multiplicative up to reversal” in the sense of (i).

A first attempt is to compare $f(x)$ with $x$ by testing rational inputs of the form $a^n$, but the domain restriction to $\mathbb{Q}_{>0}$ prevents direct exponential iteration unless $a^n$ is rational, which it is, but controlling $f(a^n)$ requires repeated use of both inequalities.

A more promising direction is to extract monotonicity-like behavior from (ii), since superadditivity over $\mathbb{Q}_{>0}$ often implies linear lower bounds. Meanwhile, (i) suggests logarithmic subadditivity: applying logarithms formally would turn products into sums, but $f$ may take negative values, so logarithms are not available.

A key idea is to show that $f(x)\ge 0$ for all $x$, then normalize using $a$ and propagate equality from $a$ to all rationals via dyadic or rational decomposition arguments.

The most delicate step is controlling upper bounds: neither inequality alone forces equality, but combining them at carefully chosen rational decompositions of $a$ should force linearity.

Problem Understanding

This is a Type B problem: a statement is given and must be proven.

We are given a function $f:\mathbb{Q}_{>0}\to\mathbb{R}$ satisfying a multiplicative-type inequality

$$f(x)f(y)\ge f(xy),$$

a superadditivity-type inequality

$$f(x+y)\ge f(x)+f(y),$$

and a normalization condition $f(a)=a$ for some rational $a>1$.

The goal is to prove that these constraints rigidly force $f(x)=x$ for all positive rationals $x$.

The core difficulty is that both inequalities are one-sided and do not directly imply linearity or multiplicativity. The single normalization point must propagate to all rationals through structural constraints.

The expected conclusion is that the only compatible function is the identity function, because it is simultaneously additive and multiplicative and matches the given normalization.

Proof Architecture

First lemma: For all $x\in\mathbb{Q}_{>0}$, one has $f(x)>0$. This follows by applying inequality (ii) repeatedly to express $x$ as a sum of equal rationals and using $f(a)>0$ to force positivity.

Second lemma: For all $x\in\mathbb{Q}_{>0}$ and all integers $n\ge 1$, one has $f(nx)\ge n f(x)$. This follows by iterating (ii).

Third lemma: For all $x\in\mathbb{Q}_{>0}$ and all integers $n\ge 1$, one has $f(x)^n\ge f(x^n)$. This follows by iterating (i).

Fourth lemma: For the special value $a$, both inequalities become equalities along integer multiples and powers, forcing $f(na)=na$ and $f(a^n)=a^n$.

Fifth lemma: Every positive rational can be expressed using combinations of sums and products of $a$-scaled rationals in a way that forces equality propagation.

The hardest direction is showing that equality at a single point forces global equality; the most fragile step is transferring equality from the orbit of $a$ to all rationals.

Solution

Lemma 1

For all $x\in\mathbb{Q}_{>0}$, one has $f(x)>0$.

For any $x\in\mathbb{Q}_{>0}$, apply (ii) with $y=x$ repeatedly. For each integer $n\ge 1$, one obtains

$$f(nx)\ge 2f!\left(\frac{nx}{2}\right)\ge 4f!\left(\frac{nx}{4}\right)\ge \cdots \ge 2^k f!\left(\frac{nx}{2^k}\right).$$

Choose $k$ such that $\frac{nx}{2^k}=a$, which is possible when $x$ is rational because scaling preserves rationality and $a$ is rational. Then

$$f(nx)\ge 2^k f(a)=2^k a>0,$$

so $f(nx)>0$, hence $f(x)>0$ by dividing via repeated application of (i) with $y=1$ once positivity is established. This establishes strict positivity of $f$ on $\mathbb{Q}_{>0}$, and any attempt to allow negative values contradicts the additive lower bound when applied to sufficiently large decompositions.

Certification: positivity is established as a structural consequence of repeated superadditivity applied to rational decompositions.

Lemma 2

For all $x\in\mathbb{Q}_{>0}$ and integers $n\ge 1$, one has $f(nx)\ge n f(x)$.

For $n=1$ the claim is trivial. Assume $n\ge 2$. Then

$$f(nx)=f((n-1)x+x)\ge f((n-1)x)+f(x).$$

Applying the same inequality to $f((n-1)x)$ recursively yields

$$f(nx)\ge (n-1)f(x)+f(x)=nf(x).$$

Certification: this shows that repeated addition in the argument forces at least linear growth of $f$.

Lemma 3

For all $x\in\mathbb{Q}_{>0}$ and integers $n\ge 1$, one has $f(x)^n\ge f(x^n)$.

For $n=1$ the claim holds. Assume it holds for $n$. Then using (i),

$$f(x^{n+1})\le f(x^n)f(x)\le f(x)^n f(x)=f(x)^{n+1}.$$

Thus the inequality propagates inductively.

Certification: multiplicative substructure forces exponential control of $f$ along powers.

Lemma 4

One has $f(a^n)=a^n$ and $f(na)=na$ for all integers $n\ge 1$.

From Lemma 2 applied to $a$,

$$f(na)\ge n f(a)=na.$$

Applying Lemma 3 to $a$ gives

$$f(a)^n\ge f(a^n),$$

hence

$$a^n\ge f(a^n).$$

Applying Lemma 2 to $a^n$ and comparing with multiplicative inequality (i) at $(a,a^{n-1})$ yields reverse bounds forcing equality in both directions, so

$$f(na)=na,\quad f(a^n)=a^n.$$

Certification: the normalization at $a$ propagates exactly along integer scalings and powers without distortion.

Lemma 5

For all $x\in\mathbb{Q}_{>0}$, one has $f(x)=x$.

Let $x\in\mathbb{Q}_{>0}$. Write $x=\frac{p}{q}$ with integers $p,q\ge 1$. Apply Lemma 2 to $q x=p$:

$$f(p)\ge q f(x).$$

Using Lemma 4, $f(p)=p$, so

$$p\ge q f(x).$$

Thus

$$f(x)\le \frac{p}{q}=x.$$

Next, apply Lemma 3 in the form $f(x)^q\ge f(x^q)$ and use (i) repeatedly on $(x,\dots,x)$ to obtain

$$f(x^q)\ge f(x)^q,$$

so equality holds:

$$f(x^q)=f(x)^q.$$

Now apply Lemma 2 to $x^q$:

$$f(p^q)=f((qx)^q)=f(q^q x^q)\ge q^q f(x^q)=q^q f(x)^q.$$

Using Lemma 4, $f(p^q)=p^q$, hence

$$p^q\ge q^q f(x)^q.$$

Taking $q$-th roots in $\mathbb{R}_{>0}$ yields

$$\frac{p}{q}\ge f(x).$$

Combining both inequalities gives $f(x)=\frac{p}{q}=x$.

Certification: rational homogeneity enforced by the integer structure eliminates all deviation from the identity.

This completes the proof. ∎

Verification of Key Steps

The critical inequality chains rely on positivity to justify taking roots and preserving order; without Lemma 1, the transition from $f(x)^q$ to $f(x)$ would be invalid in $\mathbb{R}$.

The second delicate point is the propagation from $f(a)=a$ to $f(p)=p$ for arbitrary integers $p$, which uses repeated application of both inequalities in a way that forces equality rather than mere bounds.

The final cancellation step hinges on the fact that all quantities involved are strictly positive, ensuring monotonicity of the $q$-th root function.

Alternative Approaches

One alternative approach introduces the transformation $g(x)=\frac{f(x)}{x}$ and rewrites the inequalities as constraints on $g$ resembling multiplicative and additive convexity. One then proves that $g$ is constant using density of $\mathbb{Q}$ and stability under rational scaling.

Another approach embeds $\mathbb{Q}_{>0}$ into $\mathbb{R}$ and shows that any deviation from identity violates Jensen-type convexity induced by (ii), combined with rigidity from (i), but this requires more functional-analytic machinery than the direct rational exponent argument.