if-a-gt-b-gt-0-find-the-minimum-of-a-2-1-a-b-b-

Question Number 190192 by mr W last updated on 29/Mar/23

if a>b>0, find the minimum of a^2 +(1/((a−b)b))=?

$${if}\:{a}>{b}>\mathrm{0},\:{find}\:{the}\:{minimum}\:{of} \\ $$$${a}^{\mathrm{2}} +\frac{\mathrm{1}}{\left({a}−{b}\right){b}}=? \\ $$

Commented by Frix last updated on 29/Mar/23

$$\mathrm{Minimum}=\mathrm{4}\:\mathrm{at}\:{a}=\sqrt{\mathrm{2}}\wedge{b}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$

Commented by mr W last updated on 29/Mar/23

$${what}'{s}\:{your}\:{solution}? \\ $$

Answered by cortano12 last updated on 29/Mar/23

z=a^2 +(ab−b^2 )^(−1) (dz/da)= 2a−b(ab−b^2 )^(−2) (dz/db) = −(a−2b)(ab−b^2 )^(−2) (dz/da) = 0⇒2a = (b/(b^2 (a−b)^2 )) 2a = (1/(b(a−b)^2 )) (i) (dz/db) =0⇒((2b−a)/((ab−b^2 )^2 )) =0 ; a=2b (ii) ⇒4b = (1/b^3 ) ; b=(1/( (√2))) ∧ a= (√2) ⇒ z_(min) = 2 +((√2)/(((√2)−(1/2)(√2)))) = 4

$$\:\mathrm{z}=\mathrm{a}^{\mathrm{2}} +\left(\mathrm{ab}−\mathrm{b}^{\mathrm{2}} \right)^{−\mathrm{1}} \\ $$$$\:\frac{\mathrm{dz}}{\mathrm{da}}=\:\mathrm{2a}−\mathrm{b}\left(\mathrm{ab}−\mathrm{b}^{\mathrm{2}} \right)^{−\mathrm{2}} \\ $$$$\:\frac{\mathrm{dz}}{\mathrm{db}}\:=\:−\left(\mathrm{a}−\mathrm{2b}\right)\left(\mathrm{ab}−\mathrm{b}^{\mathrm{2}} \right)^{−\mathrm{2}} \\ $$$$\:\frac{\mathrm{dz}}{\mathrm{da}}\:=\:\mathrm{0}\Rightarrow\mathrm{2a}\:=\:\frac{\mathrm{b}}{\mathrm{b}^{\mathrm{2}} \left(\mathrm{a}−\mathrm{b}\right)^{\mathrm{2}} }\: \\ $$$$\:\:\mathrm{2a}\:=\:\frac{\mathrm{1}}{\mathrm{b}\left(\mathrm{a}−\mathrm{b}\right)^{\mathrm{2}} }\:\left(\mathrm{i}\right) \\ $$$$\:\frac{\mathrm{dz}}{\mathrm{db}}\:=\mathrm{0}\Rightarrow\frac{\mathrm{2b}−\mathrm{a}}{\left(\mathrm{ab}−\mathrm{b}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\mathrm{0}\:;\:\mathrm{a}=\mathrm{2b}\:\left(\mathrm{ii}\right) \\ $$$$\Rightarrow\mathrm{4b}\:=\:\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{3}} }\:;\:\mathrm{b}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\wedge\:\mathrm{a}=\:\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{z}_{\mathrm{min}} \:=\:\mathrm{2}\:+\frac{\sqrt{\mathrm{2}}}{\left(\sqrt{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}}\right)}\:=\:\mathrm{4}\: \\ $$

Commented by mr W last updated on 29/Mar/23

$${thanks}! \\ $$

Answered by witcher3 last updated on 29/Mar/23

(a−b)b≤(1/4)((a−b)+b)^2 ...AM,GM ⇒(1/(b(a−b)))≥(4/a^2 ) a^2 +(1/((a−b)b))≥a^2 +(4/a^2 )≥2(√(.a^2 .(4/a^2 )))=4...AM,GM

$$\left(\mathrm{a}−\mathrm{b}\right)\mathrm{b}\leqslant\frac{\mathrm{1}}{\mathrm{4}}\left(\left(\mathrm{a}−\mathrm{b}\right)+\mathrm{b}\right)^{\mathrm{2}} …\mathrm{AM},\mathrm{GM} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{b}\left(\mathrm{a}−\mathrm{b}\right)}\geqslant\frac{\mathrm{4}}{\mathrm{a}^{\mathrm{2}} } \\ $$$$\mathrm{a}^{\mathrm{2}} +\frac{\mathrm{1}}{\left(\mathrm{a}−\mathrm{b}\right)\mathrm{b}}\geqslant\mathrm{a}^{\mathrm{2}} +\frac{\mathrm{4}}{\mathrm{a}^{\mathrm{2}} }\geqslant\mathrm{2}\sqrt{.\mathrm{a}^{\mathrm{2}} .\frac{\mathrm{4}}{\mathrm{a}^{\mathrm{2}} }}=\mathrm{4}…\mathrm{AM},\mathrm{GM} \\ $$

Commented by mr W last updated on 29/Mar/23

$${thanks}! \\ $$

Answered by mr W last updated on 29/Mar/23

say c=a−b >0 a^2 +(1/((a−b)b)) =(c+b)^2 +(1/(cb)) =c^2 +b^2 +2bc+(1/(cb)) =c^2 +b^2 +bc+bc+(1/(4cb))+(1/(4cb))+(1/(4cb))+(1/(4cb)) ≥8((c^2 ×b^2 ×(bc)^2 ×((1/(4cb)))^4 ))^(1/8) =8×(1/2)=4 ⇒minimum=4 when c^2 =b^2 =bc=(1/(4cb)), i.e. c=b=(1/( (√2))) & a=(√2)

$${say}\:{c}={a}−{b}\:>\mathrm{0} \\ $$$${a}^{\mathrm{2}} +\frac{\mathrm{1}}{\left({a}−{b}\right){b}} \\ $$$$=\left({c}+{b}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{{cb}} \\ $$$$={c}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{bc}+\frac{\mathrm{1}}{{cb}} \\ $$$$={c}^{\mathrm{2}} +{b}^{\mathrm{2}} +{bc}+{bc}+\frac{\mathrm{1}}{\mathrm{4}{cb}}+\frac{\mathrm{1}}{\mathrm{4}{cb}}+\frac{\mathrm{1}}{\mathrm{4}{cb}}+\frac{\mathrm{1}}{\mathrm{4}{cb}} \\ $$$$\geqslant\mathrm{8}\sqrt[{\mathrm{8}}]{{c}^{\mathrm{2}} ×{b}^{\mathrm{2}} ×\left({bc}\right)^{\mathrm{2}} ×\left(\frac{\mathrm{1}}{\mathrm{4}{cb}}\right)^{\mathrm{4}} }=\mathrm{8}×\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{4} \\ $$$$\Rightarrow{minimum}=\mathrm{4} \\ $$$${when}\:{c}^{\mathrm{2}} ={b}^{\mathrm{2}} ={bc}=\frac{\mathrm{1}}{\mathrm{4}{cb}},\:{i}.{e}.\:{c}={b}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\&\:{a}=\sqrt{\mathrm{2}} \\ $$

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