Indija TS 2008

We show that $f(x)=1/x$ is the only solution. We do this in several steps.

Step 1 We show that $f(x)$ is a non-increasing function. Suppose the contrary; say $0<a<b$ implies that $f(a)<f(b)$, for some $a,b$. Then $$w=\frac{bf(b)-af(a)}{b-a}$$ is in $(0,\infty )$. It is easy to check that $w>f(b)$ using $f(b)>f(a)$. Thus $w>f(b)>f(a)$. Taking $x=a$ and $y=w-f(a)$, we get $$f(w)=af\left(1+a(w-f(a))\right).$$ Similarly, $x=b$ and $y=w-f(b)$ gives $$f(w)=bf\left(1+b(w-f(b))\right).$$ Thus $a=b$ contradicting $a<b$. We conclude that $a<b$ implies that $f(b)\le f(a)$.

Step 2 Taking $x=y=1$, we get $f(f(1)+1)=f(2)$. Similarly, putting $x=1,y=2$, we obtain $f(f(1)+2)=f(3)$; $x=2,y=1$ implies $f(f(2)+1)=2f(3)$. Thus $$\begin{array}{rl}2f(3)& =f(f(2)+1)=f(f(f(1)+1)+1)\\ & =(f(1)+1)f(1+(f(1)+1))=(f(1)+1)f(f(1)+2)=(f(1)+1)f(3).\end{array}$$ This gives $f(1)+1=2$ and hence $f(1)=1$.

Step 3 Suppose $x>1$, and put $y=1-(1/x)$. Then $$f\left(f(x)-\frac{1}{x}+1\right)=xf(1+x-1)=xf(x).$$ If $f(x)>(1/x)$, we see that $f(x)-(1/x)+1>1$ and the monotonicity of $f(x)$ gives $$f\left(f(x)-\frac{1}{x}+1\right)\le f(1)=1.$$ Thus $xf(x)\le 1$, contradicting $f(x)>(1/x)$. If $f(x)<(1/x)$, we see that $f(x)-(1/x)+1<1$, and $$f\left(f(x)-\frac{1}{x}+1\right)\ge f(1)=1,$$ and hence $$1\le f\left(f(x)-\frac{1}{x}+1\right)=xf(x)<1,$$ which again is impossible. We conclude that $f(x)=\frac{1}{x}$, for all $x>1$.

Now, take any $x>0$. Then $f(x)+1>1$, so that $$f(f(x)+1)=\frac{1}{f(x)+1}.$$ Putting $y=1$ in the given relation, we get $$f(f(x)+1)=xf(1+x)=\frac{x}{1+x};$$ we have used $1+x>1$ in the last equality. Thus we obtain $$\frac{1}{f(x)+1}=\frac{x}{1+x},$$ for all $x>0$. This gives $f(x)=1/x$ for all $x>0$.