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 1 of 3: Find f(1) = 1 and derive the pointwise quadratic for f(x)

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In plain words

Plugging in $(1,1,1,1)$ fixes $f(1)=1$ , and plugging in $(x,1,\sqrt{x},\sqrt{x})$ gives a quadratic equation for $f(x)$ whose roots are $x$ and $\dfrac{1}{x}$ .

f(1)=1, \qquad (w,x,y,z)=(x,1,\sqrt{x},\sqrt{x}) \implies \frac{f(x)^2+1}{2f(x)} = \frac{x^2+1}{2x} \implies (xf(x)-1)(f(x)-x)=0

Detailed analysis

Setting $w=x=y=z=1$ gives $\dfrac{2f(1)^2}{2f(1)} = 1$ , so $f(1) = 1$ . Next, for any $x > 0$ , setting $(w,x,y,z) = (x, 1, \sqrt{x}, \sqrt{x})$ (which satisfies $x \cdot 1 = \sqrt{x}\cdot\sqrt{x}$ ) yields $\dfrac{f(x)^2 + 1}{2f(x)} = \dfrac{x^2 + 1}{2x}$ , or $xf(x)^2 - (x^2+1)f(x) + x = (xf(x)-1)(f(x)-x) = 0$ . Hence for each $x > 0$ , either $f(x) = x$ or $f(x) = \dfrac{1}{x}$ .