Functional equation $f\Big(\frac{x+y}{2}\Big)+f\Big(\frac{2xy}{x+y}\Big)= f(x)+f(y)$ Can someone help me please with this problem?
If the function
$f:\mathbb{R}^+\rightarrow\mathbb{R}$ satisfies the equation 
$f\Big(\frac{x+y}{2}\Big)+f\Big(\frac{2xy}{x+y}\Big)=
f(x)+f(y)$,
then it satisfies also $2f(\sqrt{xy})=f(x)+f(y)$.
I tried to collect more formulas
from substitutions like $\sqrt{xy}\rightarrow x$, $x\rightarrow\frac{x+y}{2}$
but didn't succeed.
 A: The statement is equivalent to

The value of the expression $f(x) + f(y)$ remains the same if we replace $x$ and $y$
  by their arithmetic and harmonic means:
$$\text{AM}(x,y) = \frac{x+y}{2}\quad\text{ and }\quad\text{HM}(x,y) =
\frac{1}{\frac12 ( \frac{1}{x} + \frac{1}{y}) }$$

It is known that for any $x,y \in \mathbb{R}_{+}$, we have:
$$\text{AM}(x,y) \ge \text{GM}(x,y) = \sqrt{xy} \ge \text{HM}(x,y)$$
It is also known that if we repeatly apply $\text{AM}$ and $\text{HM}$ to a pair of numbers, they will converge to GM. More precisely, if we construct two sequences $(x_n), (y_n), n\in \mathbb{N}$ by:
$$(x_n, y_n) = \begin{cases} 
(\max(x,y),\;\min(x,y)), & n = 0\\
(\text{AM(x,y)},\;\;\text{HM(x,y)}), & n > 0
\end{cases}$$
we will find
$$x_0 \ge x_1 \ge x_2 \ge \cdots \ge x_n \cdots \ge \text{GM}(x,y) \ge \cdots \ge y_n \ge \cdots \ge y_2 \ge y_1 \ge y_0$$
and $\displaystyle \lim_{n\to\infty} x_n = \lim_{n\to\infty} y_n = \text{GM}(x,y) = \sqrt{xy}$. If $f$ is continuous, this will imply
$$f(x) + f(y) = f(x_1) + f(y_1 ) = \cdots = \lim_{n->\infty} f(x_n) + f(y_n) = 2f(\sqrt{xy})$$
The part that $(x_n,y_n)$ is sandwiching $\text{GM}(x,y)$ are the standard $\text{AM}, \text{GM}, \text{HM}$ stuff, I will skip their justification. Let me demonstrate why $x_n, y_n$ converges to same limit. Notice 
$$x_{n} - y_{n} 
= \frac{x_{n-1}+y_{n-1}}{2} - \frac{2 x_{n-1} y_{n-1}}{x_{n-1}+y_{n-1}}
= \frac{(x_{n-1}-y_{n-1})^2}{2(x_{n-1}+y_{n-1})}\\
= \frac12 \left|\frac{x_{n-1}-y_{n-1}}{x_{n-1}+y_{n-1}}\right|(x_{n-1} - y_{n-1})
\le \frac12 (x_{n-1} - y_{n-1})
$$
We can conclude for general $n$, $|x_n - y_n | \le 2^{-n}|x-y|$ and hence $x_n, y_n$ converges to same limit as $n \to \infty$. Since $x_n$ and $y_n$ are sandwiching $\text{GM}(x,y)$, the common limit is $\text{GM}(x,y) = \sqrt{xy}$.
A: Hints:
$\dfrac{f(x)+f(y)}{2} \ge\sqrt{f(x)f(y)}$
Here we have  $\dfrac{f(x)+f(y)}{2} =2f(\sqrt{xy})$
