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Question Number 74213 by FCB last updated on 20/Nov/19

Commented by MJS last updated on 20/Nov/19

$$\mathrm{information}\mathrm{missing}$$$$\mathrm{I}\mathrm{think}\mathrm{the}\mathrm{question}\mathrm{is}\mathrm{find}\mathrm{minimum}\mathrm{of}\mathrm{the}$$$$\mathrm{expression}$$$$\mathrm{this}\mathrm{should}\mathrm{be}\mathrm{at}a=b=c=\frac{1}{3}$$$$\mathrm{in}\mathrm{this}\mathrm{case}\mathrm{the}\mathrm{answer}\mathrm{is}\frac{69}{4}=17.25$$

Answered by mind is power last updated on 20/Nov/19

$$whatisquation?$$

Answered by mind is power last updated on 20/Nov/19

$$a=b=0,c=1$$$$=14+\frac{7}{2}=\frac{35}{2}=17,5$$$$min\frac{7+2b}{1+a}+\frac{7+2c}{1+b}+\frac{7+2a}{1+c}$$$$a+b+c=1,a,b,c⩾0$$$$x=1+a,y=1+b,z=1+c\Rightarrow x+y+z=4$$$$\iff f(x,y,z)\frac{2y+5}{x}+\frac{5+2z}{y}+\frac{5+2x}{z}=5(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})+2(\frac{y}{x}+\frac{z}{y}+\frac{x}{z})$$$$wewilluseHa-GM-ARinqualite$$$$\frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}⩽\sqrt[3]{xyz}⩽\frac{x+y+z}{3}\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}⩾\frac{9}{x+y+z}=\frac{9}{4}$$$$(\frac{y}{x}+\frac{z}{y}+\frac{x}{z})⩾3\sqrt[3]{\left(\frac{y.x.z}{x.y.z}\right)}=3$$$$\Rightarrow f(x,y,z)⩾6+\frac{9}{4}.5=6+11.25=17.25$$$$equalitiywhenx=y=z=\frac{4}{3}\iff a=b=c=\frac{1}{3}$$$$$$

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