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Question Number 196049 by CrispyXYZ last updated on 17/Aug/23

$$a,b,c>0.\mathrm{Find}\mathrm{the}\min \mathrm{value}\mathrm{of}$$$$\sum _{\mathrm{cyc}}\sqrt{\frac{a+b}{a+b+c}}.$$

Commented by Frix last updated on 17/Aug/23

$$b=pa\wedge c=qa$$$$\mathrm{\Sigma }=\frac{\sqrt{p+1}+\sqrt{p+q}+\sqrt{q+1}}{\sqrt{p+q+1}}$$$$\mathrm{\Sigma }=2;q=\frac{1}{p}\wedge p\to \mathrm{\infty }$$$$\mathrm{\Sigma }=\sqrt{6};p=q=1$$$$\Rightarrow$$$$2<\mathrm{\Sigma }⩽\sqrt{6}$$

Answered by mr W last updated on 17/Aug/23

$$S=\frac{\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}}{\sqrt{a+b+c}}$$$$=\frac{\sqrt{2}(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a})}{\sqrt{2(a+b+c)}}$$$$=\frac{\sqrt{2}(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a})}{\sqrt{(a+b)+(b+c)+(c+a)}}$$$$=\frac{\sqrt{2}(x+y+z)}{\sqrt{x^2+y^2+z^2}}(withx=\sqrt{a+b}>0...)$$$$=\frac{\sqrt{6}(1\times x+1\times y+1\times z)}{\sqrt{3}\times \sqrt{x^2+y^2+z^2}}$$$$=\sqrt{6}\cos \theta$$$$\theta =anglebetweenvectors(1,1,1)and(x,y,z)$$$$0⩽\theta <{\cos }^{-1}\sqrt{\frac{2}{3}}$$$$\Rightarrow 2<S⩽\sqrt{6}$$$$mininum{doesn}^{\prime }texist.$$

Commented by mr W last updated on 18/Aug/23

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