x, y , z ∈R^( +) and x≥y≥z and x^2 +y^( 2) +z^( 2) ≥ 2xy +2xz+2yz Find Min((x/z) )=?


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Question Number 166212 by mnjuly1970 last updated on 15/Feb/22

$x, y, z \in R^{+} a n d x ⩾ y ⩾ z$ $a n d$ $x^{2} + y^{2} + z^{2} ⩾ 2 x y + 2 x z + 2 y z$ $F i n d M i n (\frac{x}{z}) = ?$
	Answered by mahdipoor last updated on 15/Feb/22
	${ ((x^2 +y^2 +z^2 ≥2xy+2xz+2yz)),((x^2 +y^2 +z^2 >0 )) :}⇒ 2x^2 +2y^2 +2z^2 −2xy−2xz−2zy>0⇒ A=(x−y)^2 +(y−z)^2 +(z−x)^2 >0 (I) if (z−x)=0 ⇒ x=z ⇒^(x≥y≥z) x=y=z ⇒ A=0≯0 ⇒ z−x≠0 ⇒^(x≥y≥z) x>z ⇒^(z>0) (x/z)>1$
	${\begin{cases} x^{2} + y^{2} + z^{2} ⩾ 2 x y + 2 x z + 2 y z \\ x^{2} + y^{2} + z^{2} > 0 \end{cases} \Rightarrow$ $2 x^{2} + 2 y^{2} + 2 z^{2} - 2 x y - 2 x z - 2 z y > 0 \Rightarrow$ $A = {(x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2} > 0 (I)$ $i f (z - x) = 0 \Rightarrow x = z \overset{x ⩾ y ⩾ z}{\Rightarrow} x = y = z \Rightarrow$ $A = 0 ≯ 0 \Rightarrow z - x \neq 0 \overset{x ⩾ y ⩾ z}{\Rightarrow} x > z$ $\overset{z > 0}{\Rightarrow} \frac{x}{z} > 1$
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