Question Number 142138 by mathdanisur last updated on 26/May/21
$$\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}\:+\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\:+\:\sqrt{{z}^{\mathrm{2}} +\mathrm{9}}\:=\:\mathrm{10} \\ $$$${x}+{y}+{z}=? \\ $$
Commented by mathdanisur last updated on 27/May/21
$${cool}\:{sir}\:{thank}\:{you} \\ $$
Commented by mr W last updated on 27/May/21
$$\sqrt{{y}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }+\sqrt{{x}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }+\sqrt{{z}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }\geqslant\sqrt{\left({x}+{y}+{z}\right)^{\mathrm{2}} +\left(\mathrm{1}+\mathrm{2}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\left({x}+{y}+{z}\right)^{\mathrm{2}} \leqslant\mathrm{10}^{\mathrm{2}} −\mathrm{6}^{\mathrm{2}} =\mathrm{64}=\mathrm{8}^{\mathrm{2}} \\ $$$$\Rightarrow−\mathrm{8}\leqslant{x}+{y}+{z}\leqslant\mathrm{8} \\ $$
Commented by MJS_new last updated on 26/May/21
$$\mathrm{no}\:\mathrm{unique}\:\mathrm{value} \\ $$
Commented by MJS_new last updated on 26/May/21
$$\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}={a}\:\Leftrightarrow\:{x}=\pm\sqrt{{a}^{\mathrm{2}} −\mathrm{4}};\:\mathrm{2}\leqslant{a}\leqslant\mathrm{10} \\ $$$$\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}={b}\Leftrightarrow\:{y}=\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{1}};\:\mathrm{1}\leqslant{b}\leqslant\mathrm{10} \\ $$$$\sqrt{{z}^{\mathrm{2}} +\mathrm{9}}={c}\Leftrightarrow\:{z}=\pm\sqrt{{c}^{\mathrm{2}} −\mathrm{9}};\:\mathrm{3}\leqslant{c}\leqslant\mathrm{10} \\ $$$${a}+{b}+{c}=\mathrm{10} \\ $$$$\mathrm{infinite}\:\mathrm{values}\:\mathrm{for} \\ $$$${x}+{y}+{z}=\pm\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{1}}\pm\sqrt{{c}^{\mathrm{2}} −\mathrm{9}} \\ $$
Commented by MJS_new last updated on 26/May/21
$$−\mathrm{8}\leqslant{x}+{y}+{z}\leqslant\mathrm{8} \\ $$
Commented by mathdanisur last updated on 29/May/21
$${cool}\:{Sir}\:{thanks} \\ $$