Question Number 145383 by mathdanisur last updated on 04/Jul/21
$${if}\:\:{a};{b};{c}\in\mathbb{R}^{+} \\ $$$${find}\:\:\left(\sqrt[{\mathrm{3}}]{{abc}}\:+\:\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{\mathrm{2}{b}}\:+\:\frac{\mathrm{1}}{\mathrm{4}{c}}\right)_{\boldsymbol{{min}}} =\:? \\ $$
Answered by mnjuly1970 last updated on 04/Jul/21
$$\:\mathrm{A}\:\geqslant\:\sqrt[{\mathrm{3}}]{{abc}}\:+\mathrm{3}\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{8}{abc}}} \\ $$$$\:\:\:\:\:\:=\sqrt[{\mathrm{3}}]{{abc}}\:+\frac{\mathrm{3}}{\mathrm{2}}\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{abc}}}\:\overset{{am}−{gm}} {\geqslant}\mathrm{2}\:\sqrt{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\:\mathrm{A}_{\:{min}} \:=\:\mathrm{2}\:\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}\:=\sqrt{\mathrm{6}} \\ $$
Commented by mathdanisur last updated on 04/Jul/21
$${Thanks}\:{Sir},\:{answer}\:\sqrt{\mathrm{6}} \\ $$
Commented by mnjuly1970 last updated on 04/Jul/21
$$\:{thank}\:{you}\:{so}\:{much}\:{and} \\ $$$$\:{your}\:{mention}\:… \\ $$
Commented by mathdanisur last updated on 04/Jul/21
$${cool}\:{Ser},\:{thank}\:{you} \\ $$
Answered by ajfour last updated on 04/Jul/21
$${let}\:\:\frac{\mathrm{1}}{{a}}={x}\:\:,\:\:\frac{\mathrm{1}}{\mathrm{2}{b}}={y}\:,\:\frac{\mathrm{1}}{\mathrm{4}{c}}={z} \\ $$$$\Rightarrow\:{f}\left({x},{y},{z}\right)=\frac{\mathrm{1}/\mathrm{2}}{\left({xyz}\right)^{\mathrm{1}/\mathrm{3}} }+{x}+{y}+{z} \\ $$$${minimum}\:{should}\:{of}\:{course} \\ $$$${be}\:{when}\:{x}={y}={z}\:\:\left({symmetry}\right) \\ $$$$\Rightarrow\:{f}_{{min}} =\frac{\mathrm{1}}{\mathrm{2}{x}}+\mathrm{3}{x} \\ $$$$\:\:\:\:{where}\:\:\:\frac{\mathrm{1}}{\mathrm{2}{x}}=\mathrm{3}{x}\:\:\Rightarrow\:\:{x}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\:\:\:{f}_{{min}} =\:\mathrm{6}{x}\:=\:\frac{\:\:\mathrm{6}}{\:\sqrt{\mathrm{6}}}\:=\:\sqrt{\mathrm{6}}\:. \\ $$
Commented by mathdanisur last updated on 04/Jul/21
$${cool}\:{Ser},\:{thank}\:{you} \\ $$