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if-a-b-c-R-find-abc-1-3-1-a-1-2b-1-4c-min-




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

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