Question Number 214784 by efronzo1 last updated on 19/Dec/24
$$\:\:\:\downharpoonleft\underline{\:} \\ $$
Answered by A5T last updated on 19/Dec/24
$$\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\sqrt{{c}^{\mathrm{2}} +{d}^{\mathrm{2}} }\geqslant\sqrt{\left({a}+{c}\right)^{\mathrm{2}} +\left({b}+{d}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\sqrt{\mathrm{2}^{\mathrm{2}} +{x}^{\mathrm{2}} }+\sqrt{\mathrm{3}^{\mathrm{2}} +{y}^{\mathrm{2}} }\geqslant\sqrt{\left(\mathrm{2}+\mathrm{3}\right)^{\mathrm{2}} +\left({x}+{y}\right)^{\mathrm{2}} }=\mathrm{13} \\ $$
Commented by efronzo1 last updated on 20/Dec/24
$$\mathrm{by}\:\mathrm{Euclid}\:\mathrm{theorem}? \\ $$
Commented by mr W last updated on 20/Dec/24
$$\mid{AB}\mid+\mid{BC}\mid\geqslant\mid{AC}\mid \\ $$
Answered by golsendro last updated on 20/Dec/24
$$\:\:\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }\:+\:\sqrt{\left(\mathrm{12}−\mathrm{x}\right)^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} } \\ $$$$\:\:\mathrm{min}\:\mathrm{when}\:\frac{\mathrm{x}}{\mathrm{2}}\:=\:\frac{\mathrm{12}−\mathrm{x}}{\mathrm{3}}\:\Rightarrow\mathrm{x}=\frac{\mathrm{24}}{\mathrm{5}} \\ $$$$\:\:\mathrm{min}\:\mathrm{value}\:=\sqrt{\left(\frac{\mathrm{24}}{\mathrm{5}}\right)^{\mathrm{2}} +\mathrm{4}}\:+\sqrt{\left(\mathrm{12}−\frac{\mathrm{24}}{\mathrm{5}}\right)^{\mathrm{2}} +\mathrm{9}} \\ $$$$\:=\:\mathrm{13} \\ $$