Problem 4

Find all functions

f: (0,\infty) \to (0,\infty)

(so

f

is a function from the positive real numbers) such that

\dfrac{(f(w))^2 + (f(x))^2}{f(y^2) + f(z^2)} = \dfrac{w^2 + x^2}{y^2 + z^2}

for all positive real numbers

w, x, y, z

, satisfying

wx = yz

Step 2 of 3: Rule out mixing the two branches at different points

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In plain words

Testing a pair $(a,b)$ where $f(a)=\dfrac{1}{a}$ and $f(b)=b$ with $y=z=\sqrt{ab}$ forces $f(ab)$ to equal both $\dfrac{ab(a^{-2}+b^2)}{a^2+b^2}$ and one of $ab, \dfrac{1}{ab}$ , which forces $a=1$ or $b=1$ .

f(a)=\frac{1}{a},\ f(b)=b \implies f(ab) = \frac{ab(a^{-2}+b^2)}{a^2+b^2} \in \left\{ab,\ \frac{1}{ab}\right\} \implies a=1 \text{ or } b=1

Detailed analysis

Suppose $f(a) \ne a$ and $f(b) \ne \dfrac{1}{b}$ for some $a, b > 0$ ; by Step 1, $f(a) = \dfrac{1}{a}$ (with $a \ne 1$ ) and $f(b) = b$ (with $b \ne 1$ ). Setting $(w,x,y,z) = (a, b, \sqrt{ab}, \sqrt{ab})$ gives $\dfrac{a^{-2}+b^2}{2f(ab)} = \dfrac{a^2+b^2}{2ab}$ , i.e., $f(ab) = \dfrac{ab(a^{-2}+b^2)}{a^2+b^2}$ . By Step 1, $f(ab) \in \left\{ab, \dfrac{1}{ab}\right\}$ : if $f(ab) = ab$ , then $a^{-2}+b^2 = a^2+b^2$ , forcing $a = 1$ , a contradiction; if $f(ab) = \dfrac{1}{ab}$ , then $a^2b^2(a^{-2}+b^2) = a^2+b^2$ , i.e., $b^2 + a^2b^4 = a^2+b^2$ , forcing $b = 1$ , a contradiction.