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Question Number 75618 by peter frank last updated on 13/Dec/19

Commented by mind is power last updated on 13/Dec/19

$$\mathrm{use}\mathrm{x}+\mathrm{y}⩾2\sqrt{\mathrm{xy}}$$$$(\mathrm{a}+\mathrm{b})⩾2\sqrt{\mathrm{ab}},(\mathrm{a}+\mathrm{c})⩾2\sqrt{\mathrm{ac}},\mathrm{c}+\mathrm{a}⩾2\sqrt{\mathrm{ca}}$$$$\Rightarrow (\mathrm{a}+\mathrm{b})(\mathrm{b}+\mathrm{c})(\mathrm{c}+\mathrm{a})⩾2\sqrt{\mathrm{ab}}.2\sqrt{\mathrm{bc}}.2\sqrt{\mathrm{ac}}=8\mathrm{abc}$$$$\left(\mathrm{b}\right)$$$$\mathrm{by}\mathrm{cauchy}\mathrm{shwartz}$$$$\mathrm{\Sigma }{\mathrm{x}}_{\mathrm{i}}.{\mathrm{y}}_{\mathrm{i}}⩽\sqrt{\left(\mathrm{\Sigma }{\mathrm{x}}_{\mathrm{i}}^2\right)}.\sqrt{\mathrm{\Sigma }{\mathrm{y}}_{\mathrm{i}}^2}$$$$\mathrm{x}=(\mathrm{a},\mathrm{b},\mathrm{c})\mathrm{y}=(\mathrm{b},\mathrm{c},\mathrm{a}\Rightarrow$$$$\mathrm{a}\mathrm{b}+\mathrm{bc}+\mathrm{ca}⩽\sqrt{({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}).\sqrt{({\mathrm{b}}^2+{\mathrm{c}}^2+{\mathrm{a}}^2})=({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2)$$$$2\mathrm{nd}\mathrm{methode}$$$$(\mathrm{a}+\mathrm{b})(\mathrm{b}+\mathrm{c})(\mathrm{c}+\mathrm{a})$$$$=2\mathrm{abc}+{\mathrm{a}}^2\mathrm{b}+{\mathrm{a}}^2\mathrm{c}+{\mathrm{b}}^2\mathrm{a}+{\mathrm{b}}^2\mathrm{c}+{\mathrm{c}}^2\mathrm{a}+{\mathrm{c}}^2\mathrm{b}$$$$\mathrm{we}\mathrm{use}\mathrm{AM}-\mathrm{GM}\mathrm{inequality}$$$$\Rightarrow \frac{{\mathrm{a}}^2\mathrm{b}+{\mathrm{a}}^2\mathrm{c}+{\mathrm{ab}}^2+{\mathrm{ac}}^2+{\mathrm{b}}^2\mathrm{c}+{\mathrm{c}}^2\mathrm{b}}{6}⩾{({\mathrm{a}}^2\mathrm{b}.{\mathrm{a}}^2\mathrm{c}.{\mathrm{b}}^2\mathrm{a}.{\mathrm{b}}^2\mathrm{c}.{\mathrm{c}}^2\mathrm{a}.{\mathrm{c}}^2\mathrm{b})}^{\frac{1}{6}}={\left({\mathrm{a}}^6{\mathrm{b}}^6{\mathrm{c}}^6\right)}^{\frac{1}{6}}$$$$\Rightarrow {\mathrm{a}}^2\mathrm{b}+{\mathrm{a}}^2\mathrm{c}+{\mathrm{b}}^2\mathrm{c}+{\mathrm{b}}^2\mathrm{a}+{\mathrm{c}}^2\mathrm{a}+{\mathrm{c}}^2\mathrm{b}⩾6\left(\mathrm{abc}\right)$$$$\Rightarrow (\mathrm{a}+\mathrm{b})(\mathrm{a}+\mathrm{c})(\mathrm{b}+\mathrm{c})⩾2\mathrm{abc}+6\mathrm{abc}=8\mathrm{abc}$$$$2){\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2=\frac{1}{2}({\mathrm{a}}^2+{\mathrm{b}}^2)+\frac{1}{2}({\mathrm{a}}^2+{\mathrm{c}}^2)+\frac{1}{2}({\mathrm{b}}^2+{\mathrm{c}}^2)⩾\frac{1}{2}2\mathrm{ab}+\frac{1}{2}.2\mathrm{ac}+\frac{1}{2}.2\mathrm{bc}=\mathrm{ab}+\mathrm{bc}+\mathrm{ac}$$

Commented by peter frank last updated on 14/Dec/19

$$thankyouverymuch$$

Commented by mind is power last updated on 14/Dec/19

$$\mathrm{withe}\mathrm{Pleasur}$$

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