Question Number 194297 by York12 last updated on 02/Jul/23
$${Let}\:{a}\:,\:{b}\:,\:{c}\:{be}\:\:{real}\:{positive}\:{numbers}\:\&\: \\ $$$${abc}=\mathrm{1}\: \\ $$$${prove}\:{that} \\ $$$$\frac{{ab}}{{a}^{\mathrm{5}} +{b}^{\mathrm{5}} +{ab}}+\frac{{bc}}{{b}^{\mathrm{5}} +{c}^{\mathrm{5}} +{bc}}+\frac{{ac}}{{a}^{\mathrm{5}} +{c}^{\mathrm{5}} +{ac}}\leqslant\mathrm{1} \\ $$
Answered by deleteduser1 last updated on 02/Jul/23
$${a}^{\mathrm{5}} +{b}^{\mathrm{5}} +{ab}\geqslant{a}^{\mathrm{3}} {b}^{\mathrm{2}} +{b}^{\mathrm{3}} {a}^{\mathrm{2}} +{ab}={a}^{\mathrm{2}} {b}^{\mathrm{2}} \left({a}+{b}\right)+{ab} \\ $$$$\Rightarrow\Sigma\frac{{ab}}{{a}^{\mathrm{5}} +{b}^{\mathrm{5}} +{ab}}\leqslant\Sigma\frac{\mathrm{1}}{{ab}\left({a}+{b}+{c}\right)}=\Sigma\frac{{c}}{{a}+{b}+{c}}=\mathrm{1} \\ $$
Commented by York12 last updated on 02/Jul/23
$${sir}\:{what}\:{is}\:{the}\:{motivation}\:{to}\:{use}\:{the}\:{first}\:{inequality} \\ $$$$ \\ $$
Commented by deleteduser1 last updated on 02/Jul/23
$${Rearrangement}\:{inequality} \\ $$$${Let}\:\left\{{a}_{{i}} \right\},\left\{{b}_{{i}} \right\}_{\mathrm{1}\leqslant{i}\leqslant{n}} \:{be}\:{increasing}\:{sequences}\:{or}\:{both} \\ $$$${decreasing}. \\ $$$${Then}\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} {b}_{{i}} \geqslant{a}_{\mathrm{1}} {b}_{{i}_{{p}} } +{a}_{\mathrm{2}} {b}_{{i}_{{p}_{\mathrm{2}} } } +....+{a}_{{n}} {b}_{{i}_{{p}_{{n}} } } \\ $$$${where}\:{i}_{{p}_{{j}} } \:{is}\:{the}\:{jth}\:{term}\:{in}\:{any}\:{permution}\:{of}\:\left\{{n}\right\} \\ $$
Commented by York12 last updated on 02/Jul/23
$${Thanks}\:{so}\:{much}\:{sir}\:,\:{I}\:{really}\:{appreciate} \\ $$$${it} \\ $$