Question Number 134069 by john_santu last updated on 27/Feb/21 | ||
$${Given}\:{a},{b}\:{and}\:{c}\:{are}\:{real}\:{numbers}\:{and}\:{a}<{b}<{c}. \\ $$ $${If}\:\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}}\:=\:\frac{\mathrm{1}}{\mathrm{18}}\:,\:{find}\:{minimum}\:{value}\:{of}\:{a}. \\ $$ | ||
Commented bymr W last updated on 27/Feb/21 | ||
$${no}\:{minimum}\:{for}\:{a}\:{exists}. \\ $$ | ||
Commented bymr W last updated on 27/Feb/21 | ||
$${a}=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{18}}−\left(\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\right)}\:\rightarrow\mathrm{18}\:{if}\:{b},\:{c}\:\rightarrow+\infty \\ $$ $${a}=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{18}}−\left(\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\right)}\:\rightarrow−\infty\:{if}\:{b},\:{c}\:\rightarrow\mathrm{0}^{+} \\ $$ | ||
Answered by bramlexs22 last updated on 27/Feb/21 | ||
$$\:\mathrm{AM}−\mathrm{GM} \\ $$ $$\:\frac{\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}+\frac{\mathrm{1}}{\mathrm{c}}}{\mathrm{3}\:}\:\geqslant\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{abc}}} \\ $$ $$\:\frac{\left[\:\frac{\mathrm{1}}{\mathrm{18}}\:\right]}{\mathrm{3}}\:\geqslant\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{abc}}}\:\Rightarrow\:\frac{\mathrm{1}}{\mathrm{54}}\:\geqslant\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{abc}}} \\ $$ $$\frac{\mathrm{1}}{\mathrm{54}^{\mathrm{3}} }\:\geqslant\:\frac{\mathrm{1}}{\mathrm{abc}}\:,\:\mathrm{it}\:\mathrm{hold}\:\mathrm{for}\:\mathrm{a}=\mathrm{b}=\mathrm{c} \\ $$ $$\mathrm{we}\:\mathrm{get}\:\mathrm{a}\:=\:\mathrm{54}\: \\ $$ | ||
Commented bymr W last updated on 27/Feb/21 | ||
$${what}\:{do}\:{you}\:{get}\:{if}\:{b}={c}=\mathrm{1000000}? \\ $$ $${what}\:{do}\:{you}\:{get}\:{if}\:{b}={c}=\mathrm{1}/\mathrm{1000000}? \\ $$ | ||