Question Number 59273 by naka3546 last updated on 07/May/19
$${abc}\:\:=\:\:\mathrm{64} \\ $$$${a},\:{b},\:{c}\:\:\in\:\:\mathbb{R}^{+} \\ $$$${Find}\:\:{K}\:\:{that}\:\:{satisfy}\:\:{to}\:\:{the}\:\:{inequality}\:\:: \\ $$$$\:\:\:\:\:\frac{\left({a}\:+\:{b}\right)\:\sqrt{{ab}}\:\:+\:\:\left({b}\:+\:{c}\right)\:\sqrt{{bc}}\:\:+\:\:\left({c}\:+\:{a}\right)\:\sqrt{{ca}}}{\:\sqrt{{abc}}}\:\:\:\geqslant\:\:\sqrt{{a}}\:\:+\:\:\sqrt{{b}}\:\:+\:\:\sqrt{{c}}\:\:+\:\:{K}\:\:. \\ $$
Answered by tanmay last updated on 07/May/19
$$\frac{{a}+{b}}{\mathrm{2}}\geqslant\sqrt{{ab}}\: \\ $$$$\left({a}+{b}\right)\geqslant\mathrm{2}\sqrt{{ab}}\: \\ $$$$\left({a}+{b}\right)\sqrt{{ab}}\:\geqslant\mathrm{2}{ab} \\ $$$${so}\:\left({a}+{b}\right)\sqrt{{ab}}\:+\left({b}+{c}\right)\sqrt{{bc}}\:+\left({c}+{a}\right)\sqrt{{ac}}\:\geqslant\mathrm{2}\left({ab}+{bc}+{ca}\right) \\ $$$${LHS} \\ $$$$\frac{{N}_{{r}} }{{D}_{{r}} }\geqslant\frac{\mathrm{2}\left({ab}+{bc}+{ca}\right)}{\:\sqrt{{abc}}} \\ $$$$\frac{{N}_{{r}} }{{D}_{{r}} }\geqslant\frac{{ab}+{bc}+{ca}}{\mathrm{4}} \\ $$$$\frac{{a}+{b}+{c}}{\mathrm{3}}\geqslant\left({abc}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\left({a}+{b}+{c}\right)\geqslant\mathrm{12} \\ $$$$\bigstar{now}\:\frac{{ab}+{bc}+{ca}}{\mathrm{3}}\geqslant\left({ab}×{bc}×{ca}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\frac{{ab}+{bc}+{ca}}{\mathrm{3}}\geqslant\left({abc}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$\frac{{ab}+{bc}+{ca}}{\mathrm{4}}\geqslant\frac{\mathrm{3}}{\mathrm{4}}×\left(\mathrm{64}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$\frac{{ab}+{bc}+{ca}}{\mathrm{4}}\geqslant\frac{\mathrm{3}}{\mathrm{4}}×\mathrm{16} \\ $$$$\frac{{ab}+{bc}+{ca}}{\mathrm{4}}\geqslant\mathrm{12}\bigstar \\ $$$$\frac{{N}_{{r}} }{{D}_{{r}} }\geqslant\frac{{ab}+{bc}+{ca}}{\mathrm{4}} \\ $$$$\frac{{N}_{{r}} }{{D}_{{r}} }\geqslant\frac{{ab}+{bc}+{ca}}{\mathrm{4}}\geqslant\mathrm{12} \\ $$$${RHS} \\ $$$$\sqrt{{a}}\:+\sqrt{{b}}\:+\sqrt{{c}}\:+\sqrt{{k}}\: \\ $$$$\frac{\sqrt{{a}}\:+\sqrt{{b}}\:+\sqrt{{c}}\:}{\mathrm{3}}\geqslant\left(\sqrt{{a}}\:×\sqrt{{b}}\:×\sqrt{{c}}\:\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\frac{\sqrt{{a}}\:+\sqrt{{b}}\:+\sqrt{{c}}\:}{\mathrm{3}}\geqslant\left({abc}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} \\ $$$$\left(\sqrt{{a}}\:+\sqrt{{b}}\:+\sqrt{{c}}\:\right)/\mathrm{3}\:\geqslant\left(\mathrm{64}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} \:\:\:\leftarrow{correction}\:{after}\:{Naka}\:{sir}\:{detected} \\ $$$$\sqrt{{a}}\:+\sqrt{{b}}\:+\sqrt{{c}}\:\geqslant\mathrm{2}×\mathrm{3}=\mathrm{6} \\ $$$${so} \\ $$$$\frac{{N}_{{r}} }{{D}_{{r}} }\geqslant\frac{{ab}+{bc}+{ca}}{\mathrm{4}}\geqslant\mathrm{12}\geqslant\mathrm{6}+{k} \\ $$$${so}\:{when}\:\mathrm{12}=\mathrm{6}+{k}\:\:\:\:\:\:{k}=\mathrm{6} \\ $$$${when}\:\mathrm{12}\geqslant\mathrm{6}+{k} \\ $$$$\mathrm{6}\geqslant{k} \\ $$$${i}\:{have}\:{just}\:{tried}… \\ $$$$ \\ $$
Commented by naka3546 last updated on 07/May/19
$$\sqrt{{a}}\:\:+\:\:\sqrt{{b}}\:\:+\:\:\sqrt{{c}}\:\:\:\geqslant\:\:\mathrm{3}\centerdot\mathrm{2}\:\:=\:\:\mathrm{6}\:\:? \\ $$$$\frac{{N}_{{r}} }{{D}_{{r}} }\geqslant\frac{{ab}+{bc}+{ca}}{\mathrm{4}}\geqslant\mathrm{12}\geqslant\mathrm{2}+{k}\:\:\: \\ $$$${so}\:{when}\:\mathrm{12}\:=\:\mathrm{6}\:+\:{k}\:\:\:\:\:\:{k}\:=\:\mathrm{6}\:\:\: \\ $$$${K}\:\:=\:\:\mathrm{6} \\ $$
Commented by tanmay last updated on 07/May/19
$${thank}\:{you}\:{sir}\:{for}\:{your}\:{kind}\:{perusal}\:{and}\:{detection} \\ $$$${of}\:{error}… \\ $$