Solve-for-real-numbers-2x-2-3y-2-z-2-7-x-2-y-2-z-2-2-z-x-y-

Question Number 159529 by HongKing last updated on 18/Nov/21

Solve for real numbers :

{\begin{cases} {2 x}^{2} + {3 y}^{2} + z^{2} = 7 \\ x^{2} + y^{2} + z^{2} = \sqrt{2} z (x + y) \end{cases}

Answered by 1549442205PVT last updated on 18/Nov/21

A p p l y t h e i n e q u a l i t y 2 (a^{2} + b^{2}) ⩾ {(a + b)}^{2} a n d a^{2} + b^{2} ⩾ 2 a b

w e h a v e 2 \sqrt{2} z (x + y) = 2 (x^{2} + y^{2}) + 2 z^{2}

⩾ {(x + y)}^{2} + 2 z^{2} ⩾ 2 \sqrt{{(x + y)}^{2} . 2 z^{2}} = 2 \sqrt{2} z (x + y)

I t f o l l o w s t h a t {\begin{cases} x = y \\ x + y = \sqrt{2} z \end{cases} \Rightarrow 2 x = \sqrt{2 z} \Rightarrow z = x \sqrt{2}

H e n c e f r o m t h e f i r s t e q u a t i o n w e o b t a i n

5 x^{2} + 2 x^{2} = 7 \Leftrightarrow x^{2} = 1 \Leftrightarrow x = \pm 1 = y, z = \pm \sqrt{2}

Commented by HongKing last updated on 18/Nov/21

thank you so much my dear Ser cool