Problem 2

Let

\mathbb{R}^+

denote the set of positive real numbers. Find all functions

f:\mathbb{R}^+\to\mathbb{R}^+

such that for each

x\in\mathbb{R}^+

, there is exactly one

y\in\mathbb{R}^+

satisfying

xf(y)+yf(x)\le2

Step 2 of 6: Suppose

a\ne b

are friends and derive

f(a)>1/a

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In plain words

Being a friend is a symmetric relation, so if $a$ and $b$ were distinct friends, each would separately have to be its own friend as well, which quickly becomes too restrictive.

a\ne b,\quad af(b)+bf(a)>2

Detailed analysis

Suppose for contradiction that $a\ne b$ are friends of each other. Since the friend relation forces $x$ to have a friend, and by symmetry $a$ being $b$ 's unique friend and $b$ being $a$ 's unique friend already uses up their one allowed friend, $a$ cannot also be its own friend, so $af(a)+af(a)>2$ , giving $f(a)>1/a$ ; symmetrically $f(b)>1/b$ .