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Difference between revisions of "2016 APMO Problems/Problem 5"

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

Revision as of 23:12, 12 July 2021

Problem

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 $x, y, z$.

Solution

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 $f(1)=1$.


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. $\square$

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