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 1 of 4: Check that f(x) = x + c works for every c ≥ 0

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f(x)=x+c\ (c\ge0):\qquad \sqrt{\frac{x^2+(y+c)^2}{2}}\ge\frac{x+(y+c)}{2}\ge\sqrt{x(y+c)}

Detailed analysis

For $f(x)=x+c$ with $c\ge0$ , we have $f(x)>0$ on $\mathbb{R}_{>0}$ and $\tfrac{f(x)+y}{2}=\tfrac{x+f(y)}{2}$ , so the given chain of inequalities becomes the standard QM–AM–GM inequality $\sqrt{\tfrac{a^2+b^2}{2}}\ge\tfrac{a+b}{2}\ge\sqrt{ab}$ applied to the positive reals $a=x$ and $b=f(y)=y+c$ .