Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
If you are looking for continuous functions I can provide a solution. –  clark May 3 '13 at 16:39
add comment

2 Answers

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 $\text{GM}(x,y) = \sqrt{xy}$.

share|improve this answer
add comment

Hints:

$\dfrac{f(x)+f(y)}{2} \ge\sqrt{f(x)f(y)}$

Here we have $\dfrac{f(x)+f(y)}{2} =2f(\sqrt{xy})$

share|improve this answer
2  
How do you relate the inequality $\frac{f(x)+f(y)}{2}\geq \sqrt{f(x)f(y)}$ of the function values to the argument $\sqrt{xy}$? –  Albert May 3 '13 at 15:24
    
Have you heard of AM-GM inequality? The inequality crossed my mind and it possibly does help? –  user63477 May 3 '13 at 15:25
    
You shouldn't mix up an inequality for $f(x)$ and $f(y)$ with a with a search for a relation between the arguments $x$, $y$, $\frac{x+y}{2}$, $\frac{2xy}{x+y}$ and $\sqrt{xy}$. –  Albert May 3 '13 at 15:34
    
You surely can have the inequality which give bounds and thus help solve the problems.And the hint is incomplete. –  user63477 May 3 '13 at 15:38
2  
am-gm only holds for non-negative reals, and $f$ can take values in all of $\mathbb{R}$ –  lyj May 4 '13 at 3:24
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.