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Question Number 169497 by Huy last updated on 01/May/22

$$\mathrm{Determine}\mathrm{all}\mathrm{functions}f:\mathbb{R}\to \mathbb{R}\mathrm{such}\mathrm{that}$$ $$\mathrm{a}f\left(\mathrm{b}\right)+\mathrm{b}f\left(\mathrm{a}\right)=(\mathrm{a}+\mathrm{b})f\left(\sqrt{\mathrm{ab}}\right)\mathrm{\forall }\mathrm{ab}>0$$

Answered by aleks041103 last updated on 02/May/22

$$leta=x,b=x+2dx$$ $$\sqrt{ab}=\sqrt{x(x+2dx)}=x{(1+2\frac{dx}{x})}^{1/2}=x+dx$$ $$xf(x+2dx)+(x+2dx)f\left(x\right)=2(x+dx)f(x+dx)$$ $$f(x+dx)=f+f{}^{\prime }dx+\frac{f^{″}}{2}{dx}^2$$ $$f(x+2dx)=f+2f{}^{\prime }dx+2f{}^{″}{dx}^2$$ $$x(f+2f{}^{\prime }dx+2f{}^{″}{dx}^2)+xf+2fdx=$$ $$=2(x+dx)(f+f{}^{\prime }dx+\frac{f^{″}}{2}{dx}^2)$$ $$xf+{xf}^{\prime }dx+fdx+{xf}^{″}{dx}^2=$$ $$=xf+{xf}^{\prime }dx+fdx+\frac{1}{2}{xf}^{″}{dx}^2+f^{\prime }{dx}^2+\frac{1}{2}{f}^{″}{dx}^3$$ $$\Rightarrow {xf}^{″}{dx}^2=(\frac{1}{2}{xf}^{″}+f^{\prime }){dx}^2$$ $$\frac{1}{2}{xf}^{″}=f^{\prime }$$ $$let{f}^{\prime }=g$$ $$\Rightarrow \frac{dg}{dx}=\frac{2g}{x}$$ $$\Rightarrow \frac{dg}{g}=\frac{2dx}{x}$$ $$\Rightarrow d(ln\left(g\right)-2ln\left(x\right))=0$$ $$\Rightarrow g=3{cx}^2$$ $$\Rightarrow f={cx}^3+s$$ $$\Rightarrow a({cb}^3+s)+b({ca}^3+s)=(a+b)(c{\left(ab\right)}^{3/2}+s)$$ $$\Rightarrow (a+b)(a^2+b^2)=(a+b){\left(ab\right)}^{3/2}$$ $$\Rightarrow {(a^2+b^2)}^2=a^3{b}^3$$ $$whichisgenerallynottrue!$$ $$itwouldbeiffc=0$$ $$\Rightarrow f\left(x\right)=const.istheonlysolution.$$ $$$$ $$Note!$$ $$Thisappliesonlyforcontinuousfunctions.$$

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