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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

The given inequality is equivalent to ((a^2 b^2 c^2 )/(a^3 (b+c)))+((a^2 b^2 c^2 )/(b^3 (a+c)))+((a^2 b^2 c^2 )/(c^3 (a+b)))= ((b^2 c^2 )/(a(b+c)))+((a^2 c^2 )/(b(a+c)))+((a^2 b^2 )/(c(a+b))) ≥ ^(AM−GM) (((ab+bc+ca)^2 )/(2(ab+bc+ca))) = ((ab+bc+ca)/2) ≥^(cauchi) ((3^3 (√(ab.bc.ca)))/2)=(3/2) The equality occurs if and only if a=b=c=1 Hence,(1/(a^3 (b+c)))+(1/(b^3 (a+c)))+(1/(c^3 (a+b)))≥(3/2)(q.e.d)

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

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