MathLabs

Problem 4

Find all functions f:(0,∞)→(0,∞)f: (0,\infty) \to (0,\infty) (so ff is a function from the positive real numbers) such that (f(w))2+(f(x))2f(y2)+f(z2)=w2+x2y2+z2\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,zw, x, y, z, satisfying wx=yzwx = yz.
Step 3 of 3: Verify both global candidates satisfy the functional equation
Previous stepNext step
In plain words

Both f(x)=xf(x)=x and f(x)=1xf(x)=\dfrac{1}{x} satisfy the equation identically because (wx)2=(yz)2(wx)^2 = (yz)^2 cancels the denominators in the reciprocal case.

f(x)=xorf(x)=1x:w−2+x−2y−2+z−2=(w2+x2)/(wx)2(y2+z2)/(yz)2=w2+x2y2+z2f(x)=x \quad \text{or} \quad f(x)=\frac{1}{x}: \quad \frac{w^{-2}+x^{-2}}{y^{-2}+z^{-2}} = \frac{(w^2+x^2)/(wx)^2}{(y^2+z^2)/(yz)^2} = \frac{w^2+x^2}{y^2+z^2}
Detailed analysis

By Step 2, either f(x)=xf(x) = x for all x>0x > 0 or f(x)=1xf(x) = \dfrac{1}{x} for all x>0x > 0. Checking both: f(x)=xf(x) = x makes both sides of the given equation identically equal, and f(x)=1xf(x) = \dfrac{1}{x} gives w−2+x−2y−2+z−2=w2+x2y2+z2⋅(yz)2(wx)2=w2+x2y2+z2\dfrac{w^{-2}+x^{-2}}{y^{-2}+z^{-2}} = \dfrac{w^2+x^2}{y^2+z^2} \cdot \dfrac{(yz)^2}{(wx)^2} = \dfrac{w^2+x^2}{y^2+z^2} whenever wx=yzwx = yz. Thus the solutions are f(x)=xf(x) = x for all x>0x > 0 and f(x)=1xf(x) = \dfrac{1}{x} for all x>0x > 0.