We're interested in solving $\displaystyle f(x)=x^2-a=0$ to get a value for $\displaystyle \sqrt a$, using Newton's iteration $\displaystyle x_{n+1}=\frac{1}{2} \left ( x_n+ \frac{a}{x_n} \right )$.

I must show that for any $\displaystyle x_0>0$, $\displaystyle x_n \geq \sqrt a$ holds.

I've tried algebraically to show the inequality but I failed in every attempts.

For instance if I can show that $\displaystyle x_{n-1} \left ( \frac{x_{n-1}}{2} -\sqrt a \right ) + a \geq 0$ then I'd be done.

Or equivalently if $\displaystyle \frac{1}{4} \left ( x_{n-1}^2 +\frac{2a}{x_{n-1}} + \frac{a^2}{x_{n-1}^2} \right ) \geq a$.

I have a strong feeling that I should use a well known inequality to show this straightforwardly, but I don't know which one.

I'd like some help. Thanks in advance.