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Question Number 105380 by Skabetix last updated on 28/Jul/20

Commented by Skabetix last updated on 28/Jul/20

$$plsneedhelp$$

Answered by 1549442205PVT last updated on 28/Jul/20

$$\mathrm{Apply}{\mathrm{Cauchy}}^{\prime }\mathrm{s}\mathrm{inequality}\mathrm{for}\mathrm{three}$$$$\mathrm{positive}\mathrm{numners}\mathrm{we}\mathrm{get}:$$$${\mathrm{a}}^2+\mathrm{ab}+{\mathrm{b}}^2⩾3{}^3\sqrt{{\mathrm{a}}^2.\mathrm{ab}.{\mathrm{b}}^2}=3\mathrm{ab}\Rightarrow {\log }_{\mathrm{c}}({\mathrm{a}}^2+\mathrm{ab}+{\mathrm{b}}^2)⩾{\log }_{\mathrm{c}}\left(3\mathrm{ab}\right)={\log }_{\mathrm{c}}3+{\log }_{\mathrm{c}}\mathrm{a}+{\log }_{\mathrm{c}}\mathrm{b}$$$$\mathrm{Similarly},{\log }_{\mathrm{a}}({\mathrm{b}}^2+\mathrm{bc}+{\mathrm{c}}^2)⩾{\log }_{\mathrm{a}}3+{\log }_{\mathrm{a}}\mathrm{b}+{\log }_{\mathrm{a}}\mathrm{c}$$$${\log }_{\mathrm{b}}({\mathrm{c}}^2+\mathrm{ca}+{\mathrm{a}}^2)⩾{\log }_{\mathrm{b}}3+{\log }_{\mathrm{b}}\mathrm{c}+{\log }_{\mathrm{b}}\mathrm{a}$$$$\mathrm{Adding}\mathrm{up}\mathrm{three}\mathrm{above}\mathrm{inequalities}$$$$\mathrm{we}\mathrm{get}:$$$$\mathrm{LHS}⩾{\log }_{\mathrm{a}}3+{\log }_{\mathrm{b}}3+{\log }_{\mathrm{c}}3+({\log }_{\mathrm{a}}\mathrm{b}+{\log }_{\mathrm{b}}\mathrm{a})+({\log }_{\mathrm{b}}\mathrm{c}+{\log }_{\mathrm{c}}\mathrm{b})+({\log }_{\mathrm{c}}\mathrm{a}+{\log }_{\mathrm{a}}\mathrm{c})$$$$⩾{\log }_{\mathrm{a}}3+{\log }_{\mathrm{b}}3+{\log }_{\mathrm{c}}3+6\left(1\right)(\mathrm{since}{\log }_{\mathrm{x}}\mathrm{y}+{\log }_{\mathrm{y}}\mathrm{x}⩾2\sqrt{{\log }_{\mathrm{x}}\mathrm{y}.{\log }_{\mathrm{y}}\mathrm{x}}=2\mathrm{due}\mathrm{to}{\log }_{\mathrm{x}}\mathrm{y}.{\log }_{\mathrm{y}}\mathrm{x}=1)$$$$\mathrm{From}\mathrm{the}\mathrm{hypothesis}\mathrm{a}.\mathrm{b}.\mathrm{c}=3\mathrm{we}\mathrm{get}$$$$1={\log }_{3}3={\log }_3(\mathrm{a}.\mathrm{b}.\mathrm{c})={\log }_3\mathrm{a}+{\log }_3\mathrm{b}+{\log }_3\mathrm{c}$$$$\mathrm{Therefeore},\mathrm{apply}\mathrm{the}\mathrm{inequality}(\mathrm{x}+\mathrm{y}+\mathrm{z})(\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}})⩾9(\mathrm{x},\mathrm{y},\mathrm{z}>0)\mathrm{we}\mathrm{get}:$$$${\log }_{\mathrm{a}}3+{\log }_{\mathrm{b}}3+{\log }_{\mathrm{c}}3⩾\frac{9}{{\log }_3\mathrm{a}+{\log }_3\mathrm{b}+{\log }_3\mathrm{c}}=9\left(2\right)$$$$\mathrm{From}\mathrm{the}\mathrm{inequalities}\left(1\right)\mathrm{and}\left(2\right)\mathrm{we}\mathrm{obtain}:$$$$\mathrm{LHS}⩾9+6=15(\mathrm{q}.\mathrm{e}.\mathrm{d})$$$$\mathrm{The}\mathrm{equality}\mathrm{ocurrs}\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}$$$$\mathbf{a}=\mathbf{b}=\mathbf{c}={}^3\sqrt{3}$$

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