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Question Number 151554 by mathdanisur last updated on 21/Aug/21

Answered by MJS_new last updated on 21/Aug/21

$$f(x,y)=\frac{x(x+y)}{4x^2+xy+y^2}$$$$\frac{d}{dy}\left[f(x,y)\right]=0$$$$\frac{x(3x^2-2xy-y^2)}{{(4x^2+xy+y^2)}^2}=0$$$$x\in {\mathbb{R}}^{+}\iff x>0$$$$\Rightarrow y=-3x\vee y=x\mathrm{but}x>0\wedge y>0\Rightarrow y=x$$$$\frac{d^2}{{dy}^2}\left[f(x,x)\right]=-\frac{1}{9x^2}<0\Rightarrow \mathrm{maximum}\mathrm{at}y=x$$$$f(x,x)=\frac{1}{3}=\max \left(f(x,y)\right)$$$$\Rightarrow \max (f(x,y)+f(y,z)+f(z,x))=1$$

Commented by mathdanisur last updated on 22/Aug/21

$$\mathrm{Thank}\mathrm{you}\mathbf{S}\mathrm{er}$$

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