2016 APMO Problems/Problem 5

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Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that \[(z + 1)f(x + y) = f(xf(z) + y) + f(yf(z) + x),\]for all positive real numbers $x, y, z$.


We claim that $f(x)=x$ is the only solution. It is easy to check that it works. Now, we will break things down in several claims. Let $P(x.y,z)$ be the assertion to the Functional Equation.

Claim 1: $f$ is injective.

Proof: Assume $f(a)=f(b)$ for some $a,b \in \mathbb{R}^+$. Now, from $P(x,y,a)$ and $P(x,y,b)$ we have:

\[(a+1)f(x+y)=f(xf(a)+y)+f(yf(a)+x)\] \[(b+1)f(x+y)=f(xf(b)+y)+f(yf(b)+x)\]

Now comparing, we have $a=b$ as desired. $\square$

This gives us the power to compute $f(1)$. From $P(1,1,1)$ we get $f(f(1)+1)=f(2)$ and injectivity gives $f(1)=1$. Showing that $f$ is unbounded above is also easy as we can fix $(x,y)$ and let $z$ blow up to $\infty$ in the original Functional equation..

Claim 2: $f$ is surjective.

Proof: $P(x,x,z)$ gives \[(z+1)f(2x)=2f(x+xf(z)) \iff \frac{(z+1)f(2x)}{2}=f(x+xf(z))\]

This gives that $\text{Im}(f) \in \left(\frac{f(2x)}{2},\infty\right)$. Putting $x=\frac{1}{2}$, we get $\text{Im}(f) \in \left(\frac{1}{2},\infty\right)$. By induction, surjectivity is proved as $\lim_{m \to \infty}\frac{1}{2^m}=0$ and we are essentially done. $\square$

Now, we have claim that if $a+b=c+d$ for some $a,b,c,d \in \mathbb{R}^+$, then $f(a)+f(b)=f(c)+f(d)$. This can be achieved by putting $a=xf(z)+y$, $b=yf(z)+x$, $c=x'f(z)+y'$ and $d=y'f(z)+x'$. Let us calculate $f(a)+f(b)$


Everything is fine, but we would have show that such numbers $x,y,x',y' \in \mathbb{R}^+$ do exist. We will show this in the next lemma.

Lemma 1: There exits $x,y,x',y',a,b,c,d \in \mathbb{R}^+$ such that \begin{align*} a=xf(z)+y \\ b=yf(z)+x \\ c=x'f(z)+y' \\ d=y'f(z)+x' \end{align*} Proof: Notice that $f$ is surjective. So, we will solve for $x,y,x',y'$ in the system of linear equation. For convinience, let $f(z)=k$ for $k \in \mathbb{R}^+$ as $f$ is surjective. Solving we get: \begin{align*} x=\frac{ka-b}{k^2-1}\\ y=\frac{kb-a}{k^2-1} \\ x'=\frac{kc-d}{k^2-1} \\ y'=\frac{kd-c}{k^2-1} \end{align*} Now, you can chose $z$ such that $f(z)\ne 1$ and $x,y,x',y'$ is positive. This finishes the proof to the lemma. $\square$

Claim 3: $f$ obeys Jensen's Functional Equation,i.e., \[\frac{f(x)+f(y)}{2}=f\left(\frac{x+y}{2}\right)\] Proof:

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