Lời giải:
\[\begin{array}{l}
f:R \to R, \uparrow :\frac{{2\left( {f\left( y \right) - f\left( {\frac{{x + y}}{2}} \right)} \right)}}{{f\left( x \right) - f\left( y \right)}} = \frac{{f\left( x \right) - f\left( y \right)}}{{2\left( {f\left( {\frac{{x + y}}{2}} \right) - f\left( x \right)} \right)}}\forall x \ne y\\
\Rightarrow 4\left( {f\left( y \right) - f\left( {\frac{{x + y}}{2}} \right)} \right)\left( {f\left( {\frac{{x + y}}{2}} \right) - f\left( x \right)} \right) = {\left( {f\left( x \right) - f\left( y \right)} \right)^2}\forall x,y\left( 1 \right)
\end{array}\]
Nhận xét:
Nếu $f$ thỏa đề thì $f+c$ cũng thỏa đề ($c \in R$). Nên ta có thể giả sử $f(0)=0$.
Trong (1), thay $y$ bởi $x+2z$, ta có:
\[\begin{array}{l}
4\left( {f\left( {x + 2z} \right) - f\left( {x + z} \right)} \right)\left( {f\left( {x + z} \right) - f\left( x \right)} \right) = {\left( {f\left( {x + 2z} \right) - f\left( x \right)} \right)^2} = {\left[ {\left( {f\left( {x + 2z} \right) - f\left( {x + z} \right)} \right) + \left( {f\left( {x + z} \right) - f\left( x \right)} \right)} \right]^2}\\
\Leftrightarrow {\left\{ {\left[ {f\left( {x + 2z} \right) - f\left( {x + z} \right)} \right] - \left[ {f\left( {x + z} \right) - f\left( x \right)} \right]} \right\}^2} = 0 \\ \Leftrightarrow f\left( {x + 2z} \right) - f\left( {x + z} \right) = f\left( {x + z} \right) - f\left( x \right)\\
\Leftrightarrow f\left( {x + 2z} \right) + f\left( x \right) = 2f\left( {x + z} \right),\left( 2 \right)
\end{array}\]
Trong (2), cho $x=0$, ta có:\[f\left( {2z} \right) = 2f\left( z \right)\forall z \Rightarrow f\left( {x + 2z} \right) + f\left( x \right) = f\left( {2x + 2z} \right),\left( 3 \right)\]
Trong (3), thay $z$ bởi $\dfrac{y-x}{2}$, ta có:\[f\left( x \right) + f\left( y \right) = f\left( {x + y} \right)\]
Mà $f$ lại tăng ngặt nên $f(x)=ax$ với $a>0$.
Kết luận: $f(x)=ax+b$, với $a \in R^+, b \in R$.