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Question Number 189803 by uchihayahia last updated on 22/Mar/23

what′s the minimum value of a+(1/(b(a−b))) where a>b>0 a,b∈R

$$ \\ $$$${what}'{s}\:{the}\:{minimum}\:{value}\:{of} \\ $$$${a}+\frac{\mathrm{1}}{{b}\left({a}−{b}\right)}\:{where}\:{a}>{b}>\mathrm{0}\:{a},{b}\in\mathbb{R} \\ $$$$ \\ $$

Answered by cortano12 last updated on 22/Mar/23

f(a,b)=a+b^(−1) (a−b)^(−1) =a+(ab−b^2 )^(−1) (∂f/∂a) =1−b(ab−b^2 )^(−2) =0 (∂f/∂b) =−(a−2b)(ab−b^2 )^(−2) =0 { ((1=(b/((ab−b^2 )^2 ))⇒1=(1/(b(2b−b)^2 )))),((((2b−a)/((ab−b^2 )^2 ))=0⇒a=2b)) :} ⇒b^3 = 1⇒ { ((b=1)),((a=2)) :} f(2,1)=2+(1/(1.(2−1))) = 3

$$\:\mathrm{f}\left({a},{b}\right)={a}+{b}^{−\mathrm{1}} \left({a}−{b}\right)^{−\mathrm{1}} ={a}+\left({ab}−{b}^{\mathrm{2}} \right)^{−\mathrm{1}} \\ $$$$\frac{\partial\mathrm{f}}{\partial{a}}\:=\mathrm{1}−{b}\left({ab}−{b}^{\mathrm{2}} \right)^{−\mathrm{2}} =\mathrm{0} \\ $$$$\:\frac{\partial\mathrm{f}}{\partial{b}}\:=−\left({a}−\mathrm{2}{b}\right)\left({ab}−{b}^{\mathrm{2}} \right)^{−\mathrm{2}} =\mathrm{0} \\ $$$$\:\begin{cases}{\mathrm{1}=\frac{{b}}{\left({ab}−{b}^{\mathrm{2}} \right)^{\mathrm{2}} }\Rightarrow\mathrm{1}=\frac{\mathrm{1}}{{b}\left(\mathrm{2}{b}−{b}\right)^{\mathrm{2}} }}\\{\frac{\mathrm{2}{b}−{a}}{\left({ab}−{b}^{\mathrm{2}} \right)^{\mathrm{2}} }=\mathrm{0}\Rightarrow{a}=\mathrm{2}{b}}\end{cases} \\ $$$$\:\Rightarrow{b}^{\mathrm{3}} \:=\:\mathrm{1}\Rightarrow\begin{cases}{{b}=\mathrm{1}}\\{{a}=\mathrm{2}}\end{cases}\: \\ $$$$\:\mathrm{f}\left(\mathrm{2},\mathrm{1}\right)=\mathrm{2}+\frac{\mathrm{1}}{\mathrm{1}.\left(\mathrm{2}−\mathrm{1}\right)}\:=\:\mathrm{3} \\ $$

Commented by uchihayahia last updated on 23/Mar/23

thank you

$${thank}\:{you} \\ $$

Answered by mr W last updated on 22/Mar/23

a+(1/(b(a−b))) =a−b+b+(1/(b(a−b))) =c+b+(1/(bc)) (with c=a−b>0) ≥3((c×b×(1/(bc))))^(1/3) =3 ⇒minimum is 3

$${a}+\frac{\mathrm{1}}{{b}\left({a}−{b}\right)} \\ $$$$={a}−{b}+{b}+\frac{\mathrm{1}}{{b}\left({a}−{b}\right)} \\ $$$$={c}+{b}+\frac{\mathrm{1}}{{bc}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({with}\:{c}={a}−{b}>\mathrm{0}\right) \\ $$$$\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{{c}×{b}×\frac{\mathrm{1}}{{bc}}}=\mathrm{3} \\ $$$$\Rightarrow{minimum}\:{is}\:\mathrm{3} \\ $$

Commented by manxsol last updated on 22/Mar/23

thanks, Sir W y Sr. Cortano

$${thanks},\:{Sir}\:{W}\:{y}\:{Sr}.\:{Cortano} \\ $$

Commented by mehdee42 last updated on 22/Mar/23

Bravo .Very beautiful

$${Bravo}\:.{Very}\:{beautiful} \\ $$

Commented by uchihayahia last updated on 23/Mar/23

thanks this is what i looking for

$${thanks}\:{this}\:{is}\:{what}\:{i}\:{looking}\:{for} \\ $$

Answered by ajfour last updated on 23/Mar/23

f=((ab)/b)+(1/(ab−b^2 )) =(1/( (√b)))(((ab−b^2 )/( (√b)))+((√b)/(ab−b^2 )))+b =(2/( (√b)))+b (df/db)=−(1/(b(√b)))+1 =0 ⇒ b=1 f_(min) =2+1

$${f}=\frac{{ab}}{{b}}+\frac{\mathrm{1}}{{ab}−{b}^{\mathrm{2}} } \\ $$$$\:\:\:=\frac{\mathrm{1}}{\:\sqrt{{b}}}\left(\frac{{ab}−{b}^{\mathrm{2}} }{\:\sqrt{{b}}}+\frac{\sqrt{{b}}}{{ab}−{b}^{\mathrm{2}} }\right)+{b} \\ $$$$\:=\frac{\mathrm{2}}{\:\sqrt{{b}}}+{b} \\ $$$$\frac{{df}}{{db}}=−\frac{\mathrm{1}}{{b}\sqrt{{b}}}+\mathrm{1}\:\:\:=\mathrm{0}\:\:\Rightarrow\:\:{b}=\mathrm{1} \\ $$$${f}_{{min}} =\mathrm{2}+\mathrm{1} \\ $$

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