International Mathematical Olympiad

First we prove that the function $f(x)=1/x$ satisfies the condition of the problem statement. The AM-GM inequality gives $$\frac{x}{y}+\frac{y}{x}⩾2$$ for every $x,y>0$, with equality if and only if $x=y$. This means that, for every $x>0$, there exists a unique $y>0$ such that $$\frac{x}{y}+\frac{y}{x}⩽2$$ namely $y=x$.

Let now $f:{\mathbb{R}}_{>0}\to {\mathbb{R}}_{>0}$ be a function that satisfies the condition of the problem statement. We say that a pair of positive real numbers $(x,y)$ is good if $xf(y)+yf(x)⩽2$. Observe that if $(x,y)$ is good, then so is $(y,x)$.

Lemma 1.0. If $(x,y)$ is good, then $x=y$.

Proof. Assume that there exist positive real numbers $x\ne y$ such that $(x,y)$ is good. The uniqueness assumption says that $y$ is the unique positive real number such that $(x,y)$ is good. In particular, $(x,x)$ is not a good pair. This means that $$xf(x)+xf(x)>2$$ and thus $xf(x)>1$. Similarly, $(y,x)$ is a good pair, so $(y,y)$ is not a good pair, which implies $yf(y)>1$. We apply the AM-GM inequality to obtain $$xf(y)+yf(x)⩾2\sqrt{xf(y)\cdot yf(x)}=2\sqrt{xf(x)\cdot yf(y)}>2.$$ This is a contradiction, since $(x,y)$ is a good pair.

By assumption, for any $x>0$, there always exists a good pair containing $x$, however Lemma 1 implies that the only good pair that can contain $x$ is $(x,x)$, so $$xf(x)⩽1⟺f(x)⩽\frac{1}{x},$$ for every $x>0$.

Solution 1.1. We give an alternative way to prove that $f(x)=1/x$ assuming $f(x)⩽1/x$ for every $x>0$. Indeed, if $f(x)<1/x$ then for every $a>0$ with $f(x)<1/a<1/x$ (and there are at least two of them), we have $$af(x)+xf(a)<1+\frac{x}{a}<2.$$ Hence $(x,a)$ is a good pair for every such $a$, a contradiction. We conclude that $f(x)=1/x$.

Solution 1.2. We can also conclude from Lemma 1 and $f(x)⩽1/x$ as follows. Lemma 2. The function $f$ is decreasing. Proof. Let $y>x>0$. Lemma 1 says that $(x,y)$ is not a good pair, but $(y,y)$ is. Hence $$xf(y)+yf(x)>2⩾2yf(y)>yf(y)+xf(y),$$ where we used $y>x$ (and $f(y)>0$ ) in the last inequality. This implies that $f(x)>f(y)$, showing that $f$ is decreasing.

We now prove that $f(x)=1/x$ for all $x$. Fix a value of $x$ and note that for $y>x$ we must have $xf(x)+yf(x)>xf(y)+yf(x)>2$ (using that $f$ is decreasing for the first step), hence $f(x)>\frac{2}{x+y}$. The last inequality is true for every $y>x>0$. If we fix $x$ and look for the supremum of the expression $\frac{2}{x+y}$ over all $y>x$, we get $$f(x)⩾\frac{2}{x+x}=\frac{1}{x}.$$ Since we already know that $f(x)⩽1/x$, we conclude that $f(x)=1/x$.

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Alternative solution.

As in the first solution, we note that $f(x)=1/x$ is a solution, and we set out to prove that it is the only one. We write $g(x)$ for the unique positive real number such that $(x,g(x))$ is a good pair. In this solution, we prove Lemma 2 without assuming Lemma 1.

Lemma 2. The function $f$ is decreasing. Proof. Consider $x<y$. It holds that $yf(g(y))+g(y)f(y)⩽2$. Moreover, because $y$ is the only positive real number such that $(g(y),y)$ is a good pair and $x\ne y$, we have $xf(g(y))+g(y)f(x)>2$. Combining these two inequalities yields $$xf(g(y))+g(y)f(x)>2⩾yf(g(y))+g(y)f(y),$$ or $f(g(y))(x-y)>g(y)(f(y)-f(x))$. Because $g(y)$ and $f(g(y))$ are both positive while $x-y$ is negative, it follows that $f(y)<f(x)$, showing that $f$ is decreasing.

We now prove Lemma 1 using Lemma 2. Suppose that $x\ne y$ but $xf(y)+yf(x)⩽2$. As in the first solution, we get $xf(x)+xf(x)>2$ and $yf(y)+yf(y)>2$, which implies $xf(x)+yf(y)>2$. Now $$xf(x)+yf(y)>2⩾xf(y)+yf(x)$$ implies $(x-y)(f(x)-f(y))>0$, which contradicts the fact that $f$ is decreasing. So $y=x$ is the unique $y$ such that $(x,y)$ is a good pair, and in particular we have $f(x)⩽1/x$.

We can now conclude the proof as in any of the Solutions 1.x.

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Alternative solution.

As in the other solutions we verify that the function $f(x)=1/x$ is a solution. We first want to prove the following lemma:

Lemma 3. For all $x\in {\mathbb{R}}_{>0}$ we actually have $xf(g(x))+g(x)f(x)=2$ (that is: the inequality is actually an equality).

Proof. We proceed by contradiction: Assume there exists some number $x>0$ such that for $y=g(x)$ we have $xf(y)+yf(x)<2$. Then for any $0<ϵ<\frac{2-xf(y)-yf(x)}{2f(x)}$ we have, by uniqueness of $y$, that $xf(y+ϵ)+(y+ϵ)f(x)>2$. Therefore $$\begin{array}{rl}f(y+ϵ)& >\frac{2-(y+ϵ)f(x)}{x}=\frac{2-yf(x)-ϵf(x)}{x}\\ & >\frac{2-yf(x)-\frac{2-xf(y)-yf(x)}{2}}{x}\\ & =\frac{2-xf(y)-yf(x)}{2x}+f(y)>f(y)\end{array}\text{\hspace{1em}(1)}$$ Furthermore, for every such $ϵ$ we have $g(y+ϵ)f(y+ϵ)+(y+ϵ)f(g(y+ϵ))⩽2$ and $g(y+ϵ)f(y)+yf(g(y+ϵ))>2$ (since $y\ne y+ϵ=g(g(y+ϵ)))$. This gives us the two inequalities $$f(g(y+ϵ))⩽\frac{2-g(y+ϵ)f(y+ϵ)}{y+ϵ}\text{ and }f(g(y+ϵ))>\frac{2-g(y+ϵ)f(y)}{y}.$$ Combining these two inequalities and rearranging the terms leads to the inequality $$2ϵ<g(y+ϵ)[(y+ϵ)f(y)-yf(y+ϵ)].$$ Moreover combining with the inequality (1) we obtain $2ϵ<g(y+ϵ)\left[(y+ϵ)f(y)-y\frac{2-xf(y)-yf(x)}{2x}+f(y)\right]=g(y+ϵ)\left[ϵf(y)-y\frac{2-xf(y)-yf(x)}{2x}\right]$. We now reach the desired contradiction, since for $ϵ$ sufficiently small we have that the left hand side is positive while the right hand side is negative.

With this lemma it then follows that for all $x,y\in {\mathbb{R}}_{>0}$ we have $$xf(y)+yf(x)⩾2$$ since for $y=g(x)$ we have equality and by uniqueness for $y\ne g(x)$ the inequality is strict.

In particular for every $x\in {\mathbb{R}}_{>0}$ and for $y=x$ we have $2xf(x)⩾2$, or equivalently $f(x)⩾1/x$ for all $x\in {\mathbb{R}}_{>0}$. With this inequality we obtain for all $x\in {\mathbb{R}}_{>0}$ $$2⩾xf(g(x))+g(x)f(x)⩾\frac{x}{g(x)}+\frac{g(x)}{x}⩾2$$ where the first inequality comes from the problem statement. Consequently each of these inequalities must actually be an equality, and in particular we obtain $f(x)=1/x$ for all $x\in {\mathbb{R}}_{>0}$.

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Alternative solution.

Again, let us prove that $f(x)=1/x$ is the only solution. Let again $g(x)$ be the unique positive real number such that $(x,g(x))$ is a good pair.

Lemma 4. The function $f$ is strictly convex. Proof. Consider the function $q_s(x)=f(x)+sx$ for some real number $s$. If $f$ is not strictly convex, then there exist $u<v$ and $t\in (0,1)$ such that $$f(tu+(1-t)v)⩾tf(u)+(1-t)f(v).$$ Hence $$\begin{array}{rl}{q}_s(tu+(1-t)v)& ⩾tf(u)+(1-t)f(v)+s(tu+(1-t)v)\\ & =t{q}_s(u)+(1-t)q_s(v)\end{array}$$ Let $w=tu+(1-t)v$ and consider the case $s=f(g(w))/g(w)$. For that particular choice of $s$, the function $q_s(x)$ has a unique minimum at $x=w$. However, since $q_s(w)⩾t{q}_s(u)+(1-t)q_s(v)$, it must hold $q_s(u)⩽q_s(w)$ or $q_s(v)⩽q_s(w)$, a contradiction.

Lemma 5. The function $f$ is continuous. Proof. Since $f$ is strictly convex and defined on an open interval, it is also continuous.

As in Solution 1, we can prove that $f(x)⩽1/x$. If $f(x)<1/x$, then we consider the function $h(y)=xf(y)+yf(x)$ which is continuous. Since $h(x)<2$, there exist at least two distinct $z\ne x$ such that $h(z)<2$ giving that $(x,z)$ is good pair for both values of $z$, a contradiction. We conclude that $f(x)=1/x$ as desired.