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

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

$$ \\ $$$$\:\:\:{x},\:{y}\:,\:{z}\:\in\mathbb{R}^{\:+} \:{and}\:\:{x}\geqslant{y}\geqslant{z} \\ $$$$\:\:\:\:{and}\:\: \\ $$$$\:\:\:\:{x}^{\mathrm{2}} +{y}^{\:\mathrm{2}} +{z}^{\:\mathrm{2}} \geqslant\:\mathrm{2}{xy}\:+\mathrm{2}{xz}+\mathrm{2}{yz} \\ $$$$\:\:\:\:\:\:\mathrm{F}{ind}\:\:\:\:\:\mathrm{M}{in}\left(\frac{{x}}{{z}}\:\right)=? \\ $$

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

$$\begin{cases}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \geqslant\mathrm{2}{xy}+\mathrm{2}{xz}+\mathrm{2}{yz}}\\{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} >\mathrm{0}\:\:\:\:}\end{cases}\Rightarrow \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} +\mathrm{2}{z}^{\mathrm{2}} −\mathrm{2}{xy}−\mathrm{2}{xz}−\mathrm{2}{zy}>\mathrm{0}\Rightarrow \\ $$$${A}=\left({x}−{y}\right)^{\mathrm{2}} +\left({y}−{z}\right)^{\mathrm{2}} +\left({z}−{x}\right)^{\mathrm{2}} >\mathrm{0}\:\:\left(\mathrm{I}\right) \\ $$$${if}\:\left({z}−{x}\right)=\mathrm{0}\:\Rightarrow\:{x}={z}\:\overset{{x}\geqslant{y}\geqslant{z}} {\Rightarrow}{x}={y}={z}\:\Rightarrow \\ $$$${A}=\mathrm{0}\ngtr\mathrm{0}\:\Rightarrow\:{z}−{x}\neq\mathrm{0}\:\overset{{x}\geqslant{y}\geqslant{z}} {\Rightarrow}\:{x}>{z} \\ $$$$\overset{{z}>\mathrm{0}} {\Rightarrow}\:\frac{{x}}{{z}}>\mathrm{1}\:\: \\ $$

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