Litlle 1000t wrote: Find all function $ f: R^ + \rightarrow R^ +$ such that for any $ x,y,z \in R^ +$ such that $ x + y \ge z$ , $ f(x + y - z) + f(2\sqrt {xz}) + f(2\sqrt {yz}) = f(x + y + z)$ Lemma The first we prove that for any $ a,b,c > 0$ then exist $ (x,y,z)\in R^{ + }$ such that $ a = x + y - z$ (1) $ b = 2\sqrt {xz}$ $ c = 2.\sqrt {yz}$ From equation we have $ \frac {x}{y} = \frac {b^2}{c^2}$ $ z = \frac {b^2}{4x}$ Replace to (1) we have $ a = x(\frac {c^2}{b^2} + 1) - \frac {b^2}{4x}$ $ \Leftrightarrow 4x^2.(\frac {c^2}{b^2} + 1) - 4ax - b^2 = 0$ Easy to check that this equation always have a positive solution . So our lemma claim . From the lemma we have $ a^2 + b^2 + c^2 = (x + y + z)^2$ So for any $ a,b,c > 0$ then $ f(a) + f(b) + f(c) = f(\sqrt {a^2 + b^2 + c^2})$ Consider $ G(x) = f(\sqrt {x})$ Then $ G(x) + G(y) + G(z) = G(x + y + z)$ This is Cauchy function .

I thought the same and initially thought that every solution to the Cauchy equation was a solution here, but it's not the case: the pathological solutions take negative values, so here we have only the solution $ f(x) = x^2$.

t0rajir0u wrote: I thought the same and initially thought that every solution to the Cauchy equation was a solution here, but it's not the case: the pathological solutions take negative values, so here we have only the solution $ f(x) = x^2$. $ f(x) = a x^2$($ %Error. "arbitray" is a bad command. a \in R^+$) is a solution.

Is $ f$ continious?

It need not be stipulated whether or not $ f$ is continuous; that $ f$ is nonnegative is enough to prove $ f = ax^2$ is the only family of solutions, as t0rajir0u pointed out (the graph of any solution of Cauchy's equation other than $ f(x) = ax$ is dense in $ \mathbb{R}^2$).

Solution is $ g(x)=ax$ It is easy to check that ${ \{g(x)}$ is increasing on $ R^{+}$ Induction we have $ (nx)=nf(x)$ so $ g(x)=g(1)x,\forall x\in Q$ Because $ Q$ is dense on $ R^{+}$ so for each $ x_o\in R$ exist two sequence ${ \{a_n}$ increasing and $ \{b_n\}$ decreasing such that $ \lim_{n\to\+\infty}a_n=\lim_{n\to +\infty}b_n=x_o$ Because g is increasing so $ g(b_n)\geq g(x_0)\geq g(a_n)$ It give that $ g(x_o)=ax_0$ Problem claim.

Quote: Consider G(x) = f(\sqrt {x}) Then G(x) + G(y) + G(z) = G(x + y + z) This is Cauchy function . I don't understand why G(x)=f(a)

fx+fy-fx+y-z

if we take \[ F(x)=f(\sqrt{x}) \] we get cauchy function