Menu Close

let-a-b-c-R-determine-the-minimum-value-3a-b-c-4b-a-c-5c-a-b-




Question Number 99117 by  M±th+et+s last updated on 18/Jun/20
let a,b,c ∈R determine the minimum  value    ((3a)/(b+c))+((4b)/(a+c))+((5c)/(a+b))
$${let}\:{a},{b},{c}\:\in\mathbb{R}\:{determine}\:{the}\:{minimum} \\ $$$${value} \\ $$$$ \\ $$$$\frac{\mathrm{3}{a}}{{b}+{c}}+\frac{\mathrm{4}{b}}{{a}+{c}}+\frac{\mathrm{5}{c}}{{a}+{b}} \\ $$
Answered by MJS last updated on 19/Jun/20
−∞<((3a)/(b+c))+((4b)/(a+c))+((5c)/(a+b))<+∞  i.e. let a=0∧b=1  ((3a)/(b+c))+((4b)/(a+c))+((5c)/(a+b))=((5c^2 +4)/c)  lim_(c→±∞) ((5c^2 +4)/c) =±∞
$$−\infty<\frac{\mathrm{3}{a}}{{b}+{c}}+\frac{\mathrm{4}{b}}{{a}+{c}}+\frac{\mathrm{5}{c}}{{a}+{b}}<+\infty \\ $$$$\mathrm{i}.\mathrm{e}.\:\mathrm{let}\:{a}=\mathrm{0}\wedge{b}=\mathrm{1} \\ $$$$\frac{\mathrm{3}{a}}{{b}+{c}}+\frac{\mathrm{4}{b}}{{a}+{c}}+\frac{\mathrm{5}{c}}{{a}+{b}}=\frac{\mathrm{5}{c}^{\mathrm{2}} +\mathrm{4}}{{c}} \\ $$$$\underset{{c}\rightarrow\pm\infty} {\mathrm{lim}}\frac{\mathrm{5}{c}^{\mathrm{2}} +\mathrm{4}}{{c}}\:=\pm\infty \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *