Question Number 224076 by MrAjder last updated on 18/Aug/25
$$ \\ $$$${D}=\sqrt{\left({m}−\frac{{n}^{\mathrm{2}} }{\mathrm{4}}\right)^{\mathrm{2}} +\left({e}^{{m}} −{n}\right)^{\mathrm{2}} }+\frac{{n}^{\mathrm{2}} }{\mathrm{4}}\left({m},{n}\in{R}\right),{D}_{\mathrm{min}} =? \\ $$
Answered by MrAjder last updated on 18/Aug/25
$${D}\geq\frac{\left({e}^{{m}} −{m}\right)+\left(\frac{{n}^{\mathrm{2}} }{\mathrm{4}}−{n}\right)}{\:\sqrt{\mathrm{2}}}+\frac{{n}^{\mathrm{2}} }{\mathrm{4}}\:\:\left({Q}\mathrm{M}-\mathrm{AM}\right) \\ $$$$\geq\frac{\mathrm{1}+\left(\frac{{n}^{\mathrm{2}} }{\mathrm{4}}−{n}\right)}{\:\sqrt{\mathrm{2}}}+\frac{{n}^{\mathrm{2}} }{\mathrm{4}}\:\:\:\:\:\left({e}^{{x}} \geq\mathrm{1}+{x}\right) \\ $$$$=\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{8}}\left({n}−\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}\right)^{\mathrm{2}} +\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$$\geq\sqrt{\mathrm{2}}−\mathrm{1} \\ $$