International Mathematical Olympiad

Take $$w=x=y=z=1,$$ then we get $(f(1){)}^2=f(1)$, so $f(1)=1$. For any real number $t>0$, let $w=t$, $x=1$, $y=z=\sqrt{t}$, we get $$\frac{(f(t){)}^2+1}{2f(t)}=\frac{t^2+1}{2t},$$ which implies $(tf(t)-1)(f(t)-t)=0$. So, for any $t>0$, $$f(t)=t\text{or}f(t)=\frac{1}{t}.\stackrel{◯}{1}$$ Suppose there exist $b,c\in (0,+\infty )$ such that $f(b)\ne b$, $f(c)\ne \frac{1}{c}$. By ①, we get $b,c$ different from 1 and $f(b)=\frac{1}{b}$, $f(c)=c$. Take $w=b,x=c,y=z=\sqrt{bc}$, then $$\frac{\frac{1}{b^2}+c^2}{2f(bc)}=\frac{b^2+c^2}{2bc},$$ i.e. $f(bc)=\frac{c+b^2{c}^3}{b(b^2+c^2)}$. By ①, $f(bc)=bc$ or $f(bc)=\frac{1}{bc}$. If $f(bc)=bc$, then $$bc=\frac{c+b^2{c}^3}{b(b^2+c^2)},$$ which yields $b^{4}c=c,b=1$. Contradiction!

If $f(bc)=\frac{1}{bc}$, then $$\frac{1}{bc}=\frac{c+b^2{c}^3}{b(b^2+c^2)},$$ that yields $b^2{c}^4=b^2$, $c=1$. Contradiction! Therefore, only two functions: $f(x)=x,x\in (0,+\infty )$ or $f(x)=\frac{1}{x},x\in (0,+\infty )$. It is easy to verify that these two functions satisfy the given conditions.