Question Number 189803 by uchihayahia last updated on 22/Mar/23
$$ \\ $$$${what}'{s}\:{the}\:{minimum}\:{value}\:{of} \\ $$$${a}+\frac{\mathrm{1}}{{b}\left({a}−{b}\right)}\:{where}\:{a}>{b}>\mathrm{0}\:{a},{b}\in\mathbb{R} \\ $$$$ \\ $$
Answered by cortano12 last updated on 22/Mar/23
$$\:\mathrm{f}\left({a},{b}\right)={a}+{b}^{−\mathrm{1}} \left({a}−{b}\right)^{−\mathrm{1}} ={a}+\left({ab}−{b}^{\mathrm{2}} \right)^{−\mathrm{1}} \\ $$$$\frac{\partial\mathrm{f}}{\partial{a}}\:=\mathrm{1}−{b}\left({ab}−{b}^{\mathrm{2}} \right)^{−\mathrm{2}} =\mathrm{0} \\ $$$$\:\frac{\partial\mathrm{f}}{\partial{b}}\:=−\left({a}−\mathrm{2}{b}\right)\left({ab}−{b}^{\mathrm{2}} \right)^{−\mathrm{2}} =\mathrm{0} \\ $$$$\:\begin{cases}{\mathrm{1}=\frac{{b}}{\left({ab}−{b}^{\mathrm{2}} \right)^{\mathrm{2}} }\Rightarrow\mathrm{1}=\frac{\mathrm{1}}{{b}\left(\mathrm{2}{b}−{b}\right)^{\mathrm{2}} }}\\{\frac{\mathrm{2}{b}−{a}}{\left({ab}−{b}^{\mathrm{2}} \right)^{\mathrm{2}} }=\mathrm{0}\Rightarrow{a}=\mathrm{2}{b}}\end{cases} \\ $$$$\:\Rightarrow{b}^{\mathrm{3}} \:=\:\mathrm{1}\Rightarrow\begin{cases}{{b}=\mathrm{1}}\\{{a}=\mathrm{2}}\end{cases}\: \\ $$$$\:\mathrm{f}\left(\mathrm{2},\mathrm{1}\right)=\mathrm{2}+\frac{\mathrm{1}}{\mathrm{1}.\left(\mathrm{2}−\mathrm{1}\right)}\:=\:\mathrm{3} \\ $$
Commented by uchihayahia last updated on 23/Mar/23
$${thank}\:{you} \\ $$
Answered by mr W last updated on 22/Mar/23
$${a}+\frac{\mathrm{1}}{{b}\left({a}−{b}\right)} \\ $$$$={a}−{b}+{b}+\frac{\mathrm{1}}{{b}\left({a}−{b}\right)} \\ $$$$={c}+{b}+\frac{\mathrm{1}}{{bc}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({with}\:{c}={a}−{b}>\mathrm{0}\right) \\ $$$$\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{{c}×{b}×\frac{\mathrm{1}}{{bc}}}=\mathrm{3} \\ $$$$\Rightarrow{minimum}\:{is}\:\mathrm{3} \\ $$
Commented by manxsol last updated on 22/Mar/23
$${thanks},\:{Sir}\:{W}\:{y}\:{Sr}.\:{Cortano} \\ $$
Commented by mehdee42 last updated on 22/Mar/23
$${Bravo}\:.{Very}\:{beautiful} \\ $$
Commented by uchihayahia last updated on 23/Mar/23
$${thanks}\:{this}\:{is}\:{what}\:{i}\:{looking}\:{for} \\ $$
Answered by ajfour last updated on 23/Mar/23
$${f}=\frac{{ab}}{{b}}+\frac{\mathrm{1}}{{ab}−{b}^{\mathrm{2}} } \\ $$$$\:\:\:=\frac{\mathrm{1}}{\:\sqrt{{b}}}\left(\frac{{ab}−{b}^{\mathrm{2}} }{\:\sqrt{{b}}}+\frac{\sqrt{{b}}}{{ab}−{b}^{\mathrm{2}} }\right)+{b} \\ $$$$\:=\frac{\mathrm{2}}{\:\sqrt{{b}}}+{b} \\ $$$$\frac{{df}}{{db}}=−\frac{\mathrm{1}}{{b}\sqrt{{b}}}+\mathrm{1}\:\:\:=\mathrm{0}\:\:\Rightarrow\:\:{b}=\mathrm{1} \\ $$$${f}_{{min}} =\mathrm{2}+\mathrm{1} \\ $$