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Question Number 142138 by mathdanisur last updated on 26/May/21

$$\sqrt{y^2+1}+\sqrt{x^2+4}+\sqrt{z^2+9}=10$$$$x+y+z=?$$

Commented by mathdanisur last updated on 27/May/21

$$coolsirthankyou$$

Commented by mr W last updated on 27/May/21

$$\sqrt{y^2+1^2}+\sqrt{x^2+2^2}+\sqrt{z^2+3^2}⩾\sqrt{{(x+y+z)}^2+{(1+2+3)}^2}$$$$\Rightarrow {(x+y+z)}^2⩽{10}^2-6^2=64=8^2$$$$\Rightarrow -8⩽x+y+z⩽8$$

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

$$\sqrt{x^2+4}=a\iff x=\pm \sqrt{a^2-4};2⩽a⩽10$$$$\sqrt{y^2+1}=b\iff y=\pm \sqrt{b^2-1};1⩽b⩽10$$$$\sqrt{z^2+9}=c\iff z=\pm \sqrt{c^2-9};3⩽c⩽10$$$$a+b+c=10$$$$\mathrm{infinite}\mathrm{values}\mathrm{for}$$$$x+y+z=\pm \sqrt{a^2-4}\pm \sqrt{b^2-1}\pm \sqrt{c^2-9}$$

Commented by MJS_new last updated on 26/May/21

$$-8⩽x+y+z⩽8$$

Commented by mathdanisur last updated on 29/May/21

$$coolSirthanks$$

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