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Question Number 98067 by 995563401 last updated on 11/Jun/20

Answered by Farruxjano last updated on 11/Jun/20

Answered by 1549442205 last updated on 12/Jun/20

$$\mathrm{The}\mathrm{given}\mathrm{inequality}\mathrm{is}\mathrm{equivalent}\mathrm{to}$$$$\frac{{\mathrm{a}}^2{\mathrm{b}}^2{\mathrm{c}}^2}{{\mathrm{a}}^3(\mathrm{b}+\mathrm{c})}+\frac{{\mathrm{a}}^2{\mathrm{b}}^2{\mathrm{c}}^2}{{\mathrm{b}}^3(\mathrm{a}+\mathrm{c})}+\frac{{\mathrm{a}}^2{\mathrm{b}}^2{\mathrm{c}}^2}{{\mathrm{c}}^3(\mathrm{a}+\mathrm{b})}=$$$$\frac{{\mathrm{b}}^2{\mathrm{c}}^2}{\mathrm{a}(\mathrm{b}+\mathrm{c})}+\frac{{\mathrm{a}}^2{\mathrm{c}}^2}{\mathrm{b}(\mathrm{a}+\mathrm{c})}+\frac{{\mathrm{a}}^2{\mathrm{b}}^2}{\mathrm{c}(\mathrm{a}+\mathrm{b})}\stackrel{\mathrm{AM}-\mathrm{GM}}{⩾}\frac{{(\mathrm{ab}+\mathrm{bc}+\mathrm{ca})}^2}{2(\mathrm{ab}+\mathrm{bc}+\mathrm{ca})}=\frac{\mathrm{ab}+\mathrm{bc}+\mathrm{ca}}{2}\stackrel{\mathrm{cauchi}}{⩾}\frac{3^3\sqrt{\mathrm{ab}.\mathrm{bc}.\mathrm{ca}}}{2}=\frac{3}{2}$$$$\mathrm{The}\mathrm{equality}\mathrm{occurs}\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}\mathrm{a}=\mathrm{b}=\mathrm{c}=1$$$$\mathrm{Hence},\frac{1}{{\mathrm{a}}^3(\mathrm{b}+\mathrm{c})}+\frac{1}{{\mathrm{b}}^3(\mathrm{a}+\mathrm{c})}+\frac{1}{{\mathrm{c}}^3(\mathrm{a}+\mathrm{b})}⩾\frac{3}{2}(\mathrm{q}.\mathrm{e}.\mathrm{d})$$$$$$

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