For 0<x<y, show that x<[tex]\sqrt{xy}[/tex]<1/2(x+y)<y(adsbygoogle = window.adsbygoogle || []).push({});

I have no difficulty showing that x<[tex]\sqrt{xy}[/tex] and 1/2(x+y)<y. I am having difficulty with [tex]\sqrt{xy}[/tex]<1/2(x+y).

x<y

xx<xy

x[tex]^{2}[/tex]<xy

x<[tex]\sqrt{xy}[/tex]

and

x<y

x+y<y+y

x+y<2y

1/2(x+y)<y

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# Homework Help: Intro Analysis - Real Numbers - Inequality proof

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