prove that $\sqrt{x+1} \gt \sqrt{x}$ if $x \gt 1$

Question

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 $0 \lt x \lt y$

Note that the derivative of $\sqrt(x) $ is positive for all $x > 0$, i.e. $\sqrt(x) $ increases monotonuously — User123456789, Oct 11 '19 at 13:16
Possible duplicate of If $0<x<y$, then prove that $\sqrt{x} <\sqrt{y}$ and $x <\sqrt{xy} <y$ — Martin R, Oct 11 '19 at 13:18

score 2 · Accepted Answer · answered Oct 11 '19 at 13:17

2

For $y>0,x\ge0$

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

answered Oct 11 '19 at 13:17

lab bhattacharjee

274,582

score 1 · Answer 2 · answered Oct 11 '19 at 13:17

1

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.

answered Oct 11 '19 at 13:17

Anton Grudkin

2,955

score 0 · Answer 3 · answered Oct 11 '19 at 13:20

0

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

answered Oct 11 '19 at 13:20

Dr. Sonnhard Graubner

95,283

score 0 · Answer 4 · answered Oct 11 '19 at 13:20

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.

prove that $\sqrt{x+1} \gt \sqrt{x}$ if $x \gt 1$

4 Answers4