Miklos Schweitzer 2001 5: Functional Equation conditions

Prove that if the function $f$ is defined on the set of positive real numbers, its values are real, and $f$ satisfies the equation $$f\left( \frac{x+y}{2}\right) + f\left(\frac{2xy}{x+y} \right) =f(x)+f(y)$$ for all positive $x,y$, then $$2f(\sqrt{xy})=f(x)+f(y)$$ for every pair $x,y$ of positive numbers.

Source: Miklos Schweitzer Memorial Competition 2001

I can see how the repeated application of the functional equation condition upon itself forms a bound, but how can I formally prove this?

• It would be (too) easy to prove if $f$ were assumed to be continuous, which it's not. – dxiv Feb 20 '17 at 1:51
• Found a solution here : artofproblemsolving.com/community/… – Rutger Moody Mar 3 '17 at 13:50
• @RutgerMoody Your link is now inactive - which is why it's always good to formally post an answer on this forum rather than just link a solution hosted somewhere else in the comments – user574848 Mar 31 at 9:27
• @user574848 Thx, I'll keep that in mind. – Rutger Moody Mar 31 at 13:20

By repeated application of the function property for some positive reals $$a,b,c,d$$ as you suggested: \begin{align*}f(a)+f(b)+f(c)+f(d)&= f\left({a+b\over2}\right)+f\left({2ab\over a+b}\right)+f\left({c+d\over2}\right)+f\left({2cd\over c+d}\right) \\ &=f\left({{a+b\over2}+{c+d\over2}\over2}\right)+f\left({2{a+b\over2}\cdot{c+d\over2}\over{a+b\over2}+{c+d\over2}}\right)+f\left({{2ab\over a+b}+{2cd\over c+d}\over2}\right)+f\left({2{2ab\over a+b}\cdot{2cd\over c+d}\over{2ab\over a+b}+{2cd\over c+d}}\right) \\ &= f\left({a+b+c+d\over4}\right)+f\left({(a+b)(c+d)\over a+b+c+d}\right)+f\left({abc+abd+acd+bcd\over(a+b)(c+d)}\right)+f\left({4abcd\over abc+abd+acd+bcd}\right) \end{align*}Now if we 'swap' $$b$$ and $$c$$ and repeat a similar process, one finds $$f\left({(a+b)(c+d)\over a+b+c+d}\right)+f\left({abc+abd+acd+bcd\over(a+b)(c+d)}\right)=f\left({(a+c)(b+d)\over a+b+c+d}\right)+f\left({abc+abd+acd+bcd\over(a+c)(b+d)}\right)\tag{1}$$ Now substitute $$a=c$$, $$b=\frac{a^2}{d}$$ and $$t=\frac{a}{b}+\frac{b}{a}$$ so that $$\frac{(a+b)(c+d)}{ a+b+c+d}=\frac{abc+abd+acd+bcd}{(a+b)(c+d)}=a$$ $${(a+c)(b+d)\over a+b+c+d}=a\cdot{2t\over2+t}$$ and $${abc+abd+acd+bcd\over(a+c)(b+d)}=a\cdot{2+t\over2t}$$ If we substitute these results into $$(1)$$, we find $$2f(a)=f\left(a\cdot{2t\over2+t}\right)+f\left(a\cdot{2+t\over2t}\right)\tag{2}$$ It can be seen that $$t\geq 2$$ by AMGM and thus $$\frac{2t}{2+t}\geq 1$$. Hence for all pairs $$x\geq y$$ there exist suitable $$a$$ and $$t$$ so that $$a\cdot{2t\over2+t}=x,a\cdot{2+t\over2t}=y\text{ and } \sqrt{xy}=a$$ so that $$(2)$$, and thus the required property, holds.