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

Revision as of 22:57, 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. $\square$

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 give that $\text{Im}(f) \in \left(\frac{f(2x)}{2},\infty\right)$

@@ Line 21: / Line 21: @@
 <i>Proof:</i> <math>P(x,x,z)</math> gives
 <cmath>(z+1)f(2x)=2f(x+xf(z)) \iff \frac{(z+1)f(2x)}{2}=f(x+xf(z))</cmath>
+This give that <math>\text{Im}(f) \in \left(\frac{f(2x)}{2},\infty\right)</math>

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

Revision as of 22:57, 12 July 2021

Problem

Solution