In terms of Inequalities, we have covered till AM-GM inequality and Cauchy-Schwarz Inequality in class. I was thinking about applying AM-GM inequality but not sure if its right.
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Because $$\sqrt{y}-\sqrt{x}=\frac{y-x}{\sqrt{y}+\sqrt{x}}>0,$$ $$y-\sqrt{xy}=\sqrt{y}(\sqrt{y}-\sqrt{x})>0$$ and $$\sqrt{xy}-x=\sqrt{x}(\sqrt{y}-\sqrt{x})>0.$$
Michael Rozenberg
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if we have a>b and multiply it by z, the inequality sign will remain same only if z>0 so how do we show that 1/(√y + √x) >0 ? – RiRi Oct 19 '18 at 04:30
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@RiRi Because $\sqrt{x}>0$ and $\sqrt{y}>0.$ – Michael Rozenberg Oct 19 '18 at 04:32
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Ahhh okie! because it's mention that 0<x and 0<y hence √x and √y exist. Got it now thank you! – RiRi Oct 19 '18 at 04:42
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@RiRi You are welcome! – Michael Rozenberg Oct 19 '18 at 04:52
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$0<x<y$.
$(y-x)=$
$(√y-√x)(√y+√x)>0.$
Since $√y+√x >0$, we get
$√y -√x>0$, or $√y>√x$.
$x=√x√x <√x√y <√y√y =y$
Peter Szilas
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$ \sqrt{x} <\sqrt{y} \implies x = \sqrt{x} \sqrt{x} < \sqrt{x} \sqrt{y} = \sqrt{xy} $
$ \sqrt{x} <\sqrt{y} \implies \sqrt{xy} = \sqrt{x} \sqrt{y} < \sqrt{y} \sqrt{y} = y $
lhf
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