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Question Number 21319 by Tinkutara last updated on 20/Sep/17

If x, y, z are three real numbers such  that x + y + z = 4 and x^2  + y^2  + z^2  = 6,  then  (1) (2/3) ≤ x, y, z ≤ 2  (2) 0 ≤ x, y, z ≤ 2  (3) 1 ≤ x, y, z ≤ 3  (4) 2 ≤ x, y, z ≤ 3

$$\mathrm{If}\:{x},\:{y},\:{z}\:\mathrm{are}\:\mathrm{three}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{such} \\ $$$$\mathrm{that}\:{x}\:+\:{y}\:+\:{z}\:=\:\mathrm{4}\:\mathrm{and}\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{2}} \:=\:\mathrm{6}, \\ $$$$\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\frac{\mathrm{2}}{\mathrm{3}}\:\leqslant\:{x},\:{y},\:{z}\:\leqslant\:\mathrm{2} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{0}\:\leqslant\:{x},\:{y},\:{z}\:\leqslant\:\mathrm{2} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{1}\:\leqslant\:{x},\:{y},\:{z}\:\leqslant\:\mathrm{3} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{2}\:\leqslant\:{x},\:{y},\:{z}\:\leqslant\:\mathrm{3} \\ $$

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