Solution-1

Answer: The function \(f(x)=1/x\) is the only solution.

First we prove that the function \(f(x)=1/x\) satisfies the condition of the problem statement. The AM-GM inequality gives

\[\frac{x}{y}+\frac{y}{x} \geqslant 2\]

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

\[\frac{x}{y}+\frac{y}{x} \leqslant 2\]

namely \(y=x\). Let now \(f: \mathbb{R}{>0} \rightarrow \mathbb{R}{>0}\) 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 \(x f(y)+y f(x) \leqslant 2\). 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 \(x \neq y\) 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

\[x f(x) + x f(x)>2\]

and thus \(x f(x)>1\). Similarly, \((y, x)\) is a good pair, so \((y, y)\) is not a good pair, which implies \(y f(y)>1\). We apply the AM-GM inequality to obtain

\[x f(y)+y f(x) \geqslant 2 \sqrt{x f(y) \cdot y f(x)}=2 \sqrt{x f(x) \cdot y f(y)}>2\]

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

\[x f(x) \leqslant 1 \quad \Longleftrightarrow \quad f(x) \leqslant \frac{1}{x}\]

for every \(x>0\). In particular, with \(x=1/f(t)\) for \(t>0\), we obtain

\[\frac{1}{f(t)} \cdot f\left(\frac{1}{f(t)}\right) \leqslant 1\]

Hence

\[t \cdot f\left(\frac{1}{f(t)}\right) \leqslant t f(t) \leqslant 1\]

We claim that \((t, 1 / f(t))\) is a good pair for every \(t>0\). Indeed,

\[t \cdot f\left(\frac{1}{f(t)}\right)+\frac{1}{f(t)} f(t)=t \cdot f\left(\frac{1}{f(t)}\right)+1 \leqslant 2\]

Lemma 1 implies that \(t=1 / f(t) \Longleftrightarrow f(t)=1 / t\) for every \(t>0\). ### Solution-1.1 We give an alternative way to prove that \(f(x)=1/x\) assuming \(f(x) \leqslant 1 / x\) 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

\[a f(x)+x f(a)<1+\frac{x}{a}<2\]

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 \(f(x) \leqslant 1 / x\) 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

\[x f(y)+y f(x)>2 \geqslant 2 y f(y)>y f(y)+x f(y)\]

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 \(x f(x)+y f(x)>x f(y)+y f(x)>2\) (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

\[f(x) \geqslant \frac{2}{x+x}=\frac{1}{x}\]

Since we already know that \(f(x) \leqslant 1/x\), we conclude that \(f(x)=1/x\).