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This question is in relation to the specific case where $\sqrt{3} \gt \sqrt{2}$. Can we generalize this result and prove that $\sqrt{x+1} \gt \sqrt{x}$ if $x \gt 1$. Can we also prove the more general case that $\sqrt{y} \gt \sqrt{x}$ if $0 \lt x \lt y$

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For $y>0,x\ge0$

$$\sqrt y-\sqrt x=\dfrac{y-x}{\sqrt y+\sqrt x}$$ will be $>=<0$ according as $y-x>=<0$

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Let $f(x)=\sqrt{x}$ then $$ f’(x) = \frac{1}{2\sqrt{x}} > 0, $$ so $f(x)$ is monotonically increasing function (for $x>0$), which is exactly what you need to show.

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Squaring both sides we get $$x+1>x$$ or $$1>0$$ which is true.

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for non-negative numbers $y,z$ we have $$y<z \iff y^2<z^2$$ because $f(x)=x^2$ is strictly increasing in the interval $[0,\infty[$. Therefore you can square an inequality without changing the set of solutions, if the values must both be non-negative.

Since the roots are by definition non-negative , you can square always the inequality, if both sides are a root of some expression.

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