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Question Number 199923 by Rupesh123 last updated on 11/Nov/23

Answered by deleteduser1 last updated on 11/Nov/23

$$a+1+1⩾3\sqrt[3]{a}\Rightarrow \mathrm{\Sigma }\frac{a}{a+2}⩽\mathrm{\Sigma }\frac{a^{\frac{2}{3}}}{3}$$$$Powermean\Rightarrow {\left(\frac{a^2+b^2+c^2}{3}\right)}^{\frac{1}{2}}⩾{\left(\frac{a^{\frac{2}{3}}+b^{\frac{2}{3}}+c^{\frac{2}{3}}}{3}\right)}^{\frac{3}{2}}$$$$\Rightarrow \mathrm{\Sigma }{a}^{\frac{2}{3}}⩽3\Rightarrow \mathrm{\Sigma }\frac{a}{a+2}⩽\frac{3}{3}=1$$

Commented by Rupesh123 last updated on 11/Nov/23

Very elegant, sir!

Answered by witcher3 last updated on 12/Nov/23

$$\iff 3-2(\frac{1}{\mathrm{a}+2}+\frac{1}{\mathrm{b}+2}+\frac{1}{\mathrm{c}+2})⩽1$$$$\sum _{\mathrm{Cycl}}\frac{1}{\mathrm{a}+2}⩾\frac{{(1+1+1)}^2}{6+\mathrm{a}+\mathrm{b}+\mathrm{c}}=\frac{9}{6_{}+\mathrm{a}+\mathrm{b}+\mathrm{c}}$$$$\mathrm{Ar}-\mathrm{Quadratic}\mathrm{Mean}$$$$\mathrm{cauchy}\mathrm{shwartz}$$$$\Rightarrow \mid 1.\mathrm{a}+1.\mathrm{b}+1.\mathrm{c}\mid ⩽\sqrt{3.({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2)}=3$$$$\iff 6+\mathrm{a}+\mathrm{b}+\mathrm{c}⩽9$$$$\iff \frac{9}{6+\mathrm{a}+\mathrm{b}+\mathrm{c}}⩾1$$$$\Rightarrow \sum _{\mathrm{cycl}}\frac{1}{\mathrm{a}+2}⩽1$$$$\iff 3-2\left(\sum _{\mathrm{cycl}}\frac{1}{\mathrm{a}+2}\right)=\mathrm{\Sigma }\frac{\mathrm{a}}{\mathrm{a}+2}⩽3-2=1$$$$$$

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