Problem 5

Let

\mathbb{R}_{>0}

be the set of positive real numbers. Determine all functions

f:\mathbb{R}_{>0}\to\mathbb{R}_{>0}

such that

\sqrt{\tfrac{x^2+f(y)^2}{2}}\ge\tfrac{f(x)+y}{2}\ge\sqrt{xf(y)}

for every

x,y\in\mathbb{R}_{>0}

Step 3 of 4: All positive step sizes must be equal

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\frac{(x+(m+1)A)+(y+nB)}{2}\ge\sqrt{(x+mA)(y+(n+1)B)}\ \Longleftrightarrow\ \bigl(\sqrt{x+mA}-\sqrt{y+(n+1)B}\bigr)^2\ge B-A

Detailed analysis

Suppose $x,y>0$ have $A:=d(x)>0$ and $B:=d(y)>0$ . Applying the right-hand inequality $\tfrac{f(u)+v}{2}\ge\sqrt{uf(v)}$ to $u=f^m(x)=x+mA$ and $v=f^n(y)=y+nB$ (using $f(u)=x+(m+1)A$ and $f(v)=y+(n+1)B$ ) and rearranging gives $\bigl(\sqrt{x+mA}-\sqrt{y+(n+1)B}\bigr)^2\ge B-A$ . If $B>A$ , then for every $m,n\ge1$ the numbers $\sqrt{x+mA}$ and $\sqrt{y+(n+1)B}$ must stay at distance at least $\sqrt{B-A}>0$ ; however, as $m\to\infty$ the sequence $\sqrt{x+mA}$ goes to $\infty$ with consecutive gaps $\sqrt{x+(m+1)A}-\sqrt{x+mA}\to0$ , so for large $n$ it passes within less than $\sqrt{B-A}$ of $\sqrt{y+(n+1)B}$ , a contradiction. Hence $B\le A$ , and by symmetry $A=B$ . Thus there is a single constant $D\ge0$ such that $f(x)\in\{x,x+D\}$ for every $x>0$ .