Question Number 180953 by depressiveshrek last updated on 19/Nov/22
$${What}'{s}\:{the}\:{smallest}\:{value}\:{of} \\ $$ $${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\frac{\mathrm{1}}{{ab}}\:{for}\:{a},\:{b}>\mathrm{0}? \\ $$
Answered by mr W last updated on 19/Nov/22
$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\frac{\mathrm{1}}{{ab}}\geqslant\mathrm{2}{ab}+\frac{\mathrm{1}}{{ab}}\geqslant\mathrm{2}\sqrt{\mathrm{2}{ab}×\frac{\mathrm{1}}{{ab}}}=\mathrm{2}\sqrt{\mathrm{2}} \\ $$ $${minimum}=\mathrm{2}\sqrt{\mathrm{2}} \\ $$ $${when}\:{a}={b}\:{and}\:\mathrm{2}{ab}=\frac{\mathrm{1}}{{ab}},\:{i}.{e}.\:{a}={b}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{2}}} \\ $$
Answered by mr W last updated on 20/Nov/22
$${an}\:{other}\:{way} \\ $$ $${f}\left({a},{b}\right)={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\frac{\mathrm{1}}{{ab}} \\ $$ $$\frac{\partial{f}}{\partial{a}}=\mathrm{2}{a}−\frac{\mathrm{1}}{{a}^{\mathrm{2}} {b}}=\mathrm{0}\:\Rightarrow\mathrm{2}{a}^{\mathrm{3}} {b}=\mathrm{1}\:\:\:...\left({i}\right) \\ $$ $$\frac{\partial{f}}{\partial{b}}=\mathrm{2}{b}−\frac{\mathrm{1}}{{ab}^{\mathrm{2}} }=\mathrm{0}\:\Rightarrow\mathrm{2}{ab}^{\mathrm{3}} =\mathrm{1}\:\:\:...\left({ii}\right) \\ $$ $${from}\:\left({i}\right)\:{and}\:\left({ii}\right): \\ $$ $${a}={b}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{2}}} \\ $$ $$\Rightarrow{f}_{{min}} =\mathrm{2}\sqrt{\mathrm{2}} \\ $$ $$\left({it}'{s}\:{minimum},\:{not}\:{maximum}\right. \\ $$ $$\left.{because}\:\frac{\partial^{\mathrm{2}} {f}}{\partial{a}^{\mathrm{2}} }>\mathrm{0}\:{and}\:\frac{\partial^{\mathrm{2}} {f}}{\partial{b}^{\mathrm{2}} }>\mathrm{0}\right) \\ $$