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 3 of 6: Contradict

af(b)+bf(a)\le2

via AM–GM

Previous step Next step

af(b)+bf(a)>\tfrac{a}{b}+\tfrac{b}{a}\ge2

Detailed analysis

Using $f(a)>1/a$ and $f(b)>1/b$ from the previous step, $af(b)+bf(a)>a/b+b/a\ge2$ by AM–GM, contradicting the assumption that $a$ and $b$ are friends, i.e. $af(b)+bf(a)\le2$ . Hence no two distinct numbers can be friends of each other, so every $x\in\mathbb{R}^+$ is its own unique friend.