Proving inequality related to certain property of function

[songoku]

Homework Statement

Consider a real valued function f which satisfies the equation f (x+y) = f (x) . f (y) for all real numbers x and y. Prove:

f ((x + y) / 2) ≤ 1/2 (f(x) + f(y))

Homework Equations

Not sure

The Attempt at a Solution

Please give me a hint to start solving this question. I have found that f (x) = (f (x/2))² but I don't know what to do next.

Thanks

 

[Buffu]

Use AM-GM, ##{A+B\over 2} \ge \sqrt{AB}## where ##A,B >0##.

 

[songoku]

Use AM-GM, ##{A+B\over 2} \ge \sqrt{AB}## where ##A,B >0##.

Thank you

 

[issacnewton]

For any ##a,b\in \mathbb{R}##, we have the following ##(a-b)^2 \geqslant 0##. Then let ##a = f(\frac{x}{2})## and ##b = f(\frac{y}{2})##, and see if you reach conclusion.

 

[issacnewton]

Songoku, when you are asked to prove something, sometimes, its wise to assume the result which you have been asked to prove. And you have both, the hypothesis and the conclusion. And with it, you try to see where does this lead to. Lot of times, this leads to some other familiar result. And then you can work backwards from that familiar result. This is one of those situations. Here if you assume the result, then using the hypothesis, you reach the result ##(a-b)^2 \geqslant 0##, if you let ##a = f(\frac{x}{2})## and ##b = f(\frac{y}{2})##. But now ##(a-b)^2 \geqslant 0## is a familiar result, which is true for all ##(a-b) \in \mathbb{R}## . Now you can work backwards.

 

[songoku]

Songoku, when you are asked to prove something, sometimes, its wise to assume the result which you have been asked to prove. And you have both, the hypothesis and the conclusion. And with it, you try to see where does this lead to. Lot of times, this leads to some other familiar result. And then you can work backwards from that familiar result. This is one of those situations. Here if you assume the result, then using the hypothesis, you reach the result ##(a-b)^2 \geqslant 0##, if you let ##a = f(\frac{x}{2})## and ##b = f(\frac{y}{2})##. But now ##(a-b)^2 \geqslant 0## is a familiar result, which is true for all ##(a-b) \in \mathbb{R}## . Now you can work backwards.

Thank you very much for the advice