Rioplatense Mathematical Olympiad

The answer is $\alpha =0$. In this case, the function $f(x)=x$ satisfies the statement. From now on we assume $\alpha >0$. Note that if such a function $f$ exists, then it is strictly increasing: indeed, taking $z>y$, there is $x\in {\mathbb{R}}^{+}$ such that $z=x^{\alpha }+y$, from where we obtain: $$f(z)=f(x^{\alpha }+y)=(f(x+y){)}^{\alpha }+f(y)>f(y).$$

CLAIM 1: $f$ is unbounded. Letting $x=1$, we obtain $f(y+1)=(f(y+1){)}^{\alpha }+f(y)\ge f(1{)}^{\alpha }+f(y)$. So $f(y+1)-f(y)\ge f(1{)}^{\alpha }$, and we can prove (by telescopic summation) that $f(n)-f(1)\ge (n-1)(f(1){)}^{\alpha }$ for all $n\in \mathbb{N}$, from which we can conclude that $f$ is unbounded.

If $\alpha =1$, then clearly there is no such function $f$. Let us consider two cases: Case 1: $\alpha >1$. In this case, taking $0<x<1$, we have: $$x+y>x^{\alpha }+y\Rightarrow f(x+y)>f(x^{\alpha }+y)\Rightarrow (f(x+y){)}^{\alpha }>(f(x^{\alpha }+y){)}^{\alpha }.$$ But from the original equation we know that $f(x^{\alpha }+y)>(f(x+y){)}^{\alpha }$, whence we conclude that $f(x^{\alpha }+y)>(f(x^{\alpha }+y){)}^{\alpha }$. Making $x^{\alpha }+y=z$, we get $f(z)>(f(z){)}^{\alpha }$, for all $z\in {\mathbb{R}}^{+}$ (because every positive real can be written in the form $x^{\alpha }+y$ with $0<x<1$). As $\alpha >1$, we conclude that $f(z)<1$ for all $z\in {\mathbb{R}}^{+}$. But this contradicts the fact that $f$ is unbounded.

Case 2: $0<\alpha <1$. In this case, we take $x>1$. Then: $$x+y>x^{\alpha }+y\Rightarrow f(x+y)>f(x^{\alpha }+y)\Rightarrow (f(x+y){)}^{\alpha }>(f(x^{\alpha }+y){)}^{\alpha }.$$

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As in the previous case, since every $z>1$ can be written as $x^{\alpha }+y$ with $x>1$, we obtain $f(z)>(f(z){)}^{\alpha }$ for all $z>1$. In this case, as $0<\alpha <1$, we conclude that $f(z)>1$ for all $z>1$. CLAIM 2: For all $k\in \mathbb{N}$, if $z>1$, then $f(z)>k$. (This implies that such a function cannot exist.) The proof is by induction. We have already proved the base case $k=1$. Now, suppose that $f(z)>k$, for all $z>1$. Then, taking $y>1$ such that $z=x^{\alpha }+y$, we obtain: $$f(z)=f(x^{\alpha }+y)=(f(x+y){)}^{\alpha }+f(y)>1+k,$$ and the induction is complete. Therefore, the only possible value is $\alpha =0$.