# prove that $\sqrt{x+1} \gt \sqrt{x}$ if $x \gt 1$ [duplicate]

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 $$0 \lt x \lt y$$

For $$y>0,x\ge0$$
$$\sqrt y-\sqrt x=\dfrac{y-x}{\sqrt y+\sqrt x}$$ will be $$>=<0$$ according as $$y-x>=<0$$
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.
Squaring both sides we get $$x+1>x$$ or $$1>0$$ which is true.
for non-negative numbers $$y,z$$ we have $$y because $$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.